THE  LIBRARY 

OF 
THE  UNIVERSITY 

OF  CALIFORNIA 
LOS  ANGELES 


STATE  NORMAL  SCHOOL 

LOS  ANGELES,  CALlFOftNiA 


SOUTHERN   BRANCH 

UNIVERSITY  OF  CALIFORNfA 
ijLlBRARY 

LOS   ANGELES,  CALIF. 


PRACTICAL  METHODS 


IN 


ARITHMETIC 


Nr 


•wa 


BY 


R.  S.  GALER,  A.  M., 

Principal  of  the  Iowa  City  Academy,  Iowa  City,  Iowa. 


SECOND  EDITION,    REVISED. 

CHICAGO: 

A.   FLANAGAN, 
1892. 


COPIBI6HT,  1889,  BY  R.   S.  GALEB. 


103 


CONTENTS, 


PAGE. 

I.     INTKODUCTION   -  5 

II.     SHORT  METHODS      -  -  -  9 

III.     PRACTICAL  MEASUREMENTS — 

Gallons  and  Barrels  in  Parallelopipedon 
— Gallons  and  Barrels  in  Cylindrical 
Vessel — Gallons  and  Barrels  in  Jug 
Cistern — Grain  in  Bin  or  Wagon — 
Hay  in  Rick — Hay  in  Bound  Stack — 
Perch  of  Book  to  Wall  a  Well— Brick 
to  Wall  a  Well  or  Cistern — Quantity 
of  Fluid  that  will  flow  through  a  Pipe 
— How  High  an  Object  must  be  to  be 
Visible  a  Certain  Distance — How  Far 
an  Object,  whose  Height  is  Known, 
is  Visible — How  Far  a  Body  will  Fall 
in  a  Given  Time  -  -  16-25 

Lumber  Measure  -  25 

Shingle  Measure  26 

Lath  Measure  27 

IV.     COMMON  FRACTIONS  28 

Addition  and  Subtraction  of  Fractions,     28 
Multiplication  and  Division  of  Frac- 
tions 30 
V.     DECIMAL  FRACTIONS 

Contracted  Method  of  Multiplication 
and  Division  -  *  -  32 


CONTENTS. 

PAGE. 

VI.     PUBLIC  SURVEYS  -          35 

VII.     THE  METRIC  SYSTEM     -  40 

VIII.     LONGITUDE  AND  TIME  43 

IX.     RATIO    -  47 

X.     PROPORTION  48 

Simple  Proportion       -  49 

Compound    Proportion  —  Cause    and 

Effect  51 

XI.     GENERAL  PRINCIPLES — RULES  53 

XII.     PERCENTAGE  55 

XIII.     PERCENTAGE — APPLICATIONS  55 

Commission  and  Brokerage  -                 CO 

Profit  and  Loss     -  63 

Stocks  and  Bonds  -                 G6 

Premium  and  Discount     -  70 

Investments     -  72 

Contracts  76 

Interest  -                 78 

Promissory  Notes  82 

Partial  Payments         -  -                  85 

U.  S.  Rule  85 

Mercantile  Rule       -  - 

True  Discount      -  89 

Bank  Discount             -  -                 91 

Exchange  96 

XIV.     HORNER'S  METHOD  FOR  THE  EXTP ACTION 

OF  ROOTS  -        102 

XV.     BITS  OF  PRACTICAL  KNOWLEDGE,  105 

XVI.     PRACTICAL  PROBLEMS           -  -        110 

XVII.    ANSWERS  112 


PREFACE. 


This  work  is  not  designed  to  be  a  complete  Arithme- 
tic. In  an  experience  of  several  years  in  teaching,  the 
author  was  able  by  a  little  ingenuity  to  abridge  mathe- 
matical calculations,  and  to  perceive  that  many  princi- 
ples in  Arithmetic  have  a  general  application.  (For 
these  he  has  formulated  rules  which  have  at  once  the 
merit  of  brevity  and  scientific  accuracy.  Moreover 
these  rules  have  as  wide  an  application  as  the  princi- 
ples for  which  they  stand,  thus  greatly  simplifying 
arithmetical  operations  and  reducing  the  number  of 
rules  to  be  committed  to  memory.  In  addition  to  this 
he  has  shortened  the  work  required  for  many  practical 
measurements. 

The  constant  demand  in  the  school  room  for  these 
methods  and  formulas  'led  the  Author  to  believe  they 
might  be  acceptable  to  a  larger  class  of  students  and 
teachers  than  he  could  reach  personally. 

With  this  aim  in  view  this  little  volume  is  published- 
It  is  not  claimed  that  all  of  it  is  original.  Much  that 
was  considered  of  value  has  been  gathered  from  other 
sources.  The  Author  has  taken  this  step  with  the  pur- 
pose of  embodying  the  results  of  his  own  individual 
experience  in  a  convenient  form,  and  at  the  same  time 
of  furnishing  a  book  of  general  value  to  pupils  and 
teachers. 


PREFACE. 

Especial  attention  is  called  to  the  manner  of  treating 
Percentage  and  its  numerous  applications.  This 
method  is  entirely  original,  and,  it  is  believed,  results 
in  greatly  simplifying  these  subjects. 

With  the  hope  that  it  may  aid  many  in  saving  much 
time  and  labor,  and  stimulate  them  to  think  independ- 
ently on  practical  subjects,  this  little  volume  is  given  to 
the  public. 

BLOOMFIELP,  IOWA,  JAN.  8th,  1889. 


TO  THE  STUDENT. 


The  following  general  rules  for  work  should  be  con- 
tinually and  faithfully  observed  : 

I.  Use  the  shortest  method  that  is  consistent  with 
clearness  and  accuracy. 

II.  In  multiplying,  dividing,  etc.,  contract  wherever 
possible. 

III.  Burden  the  mind  with  few  rules,  but  learn  to 
apply  these  rules  intelligently. 

IV.  Do  not  guess  at  methods   in  the  solution  of 
problems.     Reason  out  each  step,  and  have  faith  in  the 
result. 

V.  Do  as  much  work  mentally  as  possible. 

VI.  Remember  that  Rapidity  and  Accuracy  are  the 
great  aims. 

VII.  Bear  in  mind  that  nothing  will  take  the  place 
of  Reasoning  in  mathematical  work. 


SHORT  METHODS. 


Much  time  and  woik  may  be  saved  in  arithmetical 
computations,  by  the  use  of  certain  short  methods, 
which  are  usually  of  easy  application.  A  number  of 
them  are  given  here  which  it  is  intended  that  students 
should  use  whenever  possible.  It  requires  a  quick 
perception  that  will  come  only  with  practice,  to  tell  at 
a  glance  which  one  may  be  used  to  the  greatest  advan- 
tage. 

FIRST   METHOD. 

To  multiply  where  the  left  hand  figures  are  the  same, 
and  the  sum  cf  the  two  right  hand  figures  is  ten. 

EXAMPLE. 

55 

55 


3025  Ans. 

Beginning  at  the  right  we  say,  5  times  5  are  25. 
Then  adding  1  to  the  lower  left-hand  figure  and  multi- 
plying it  by  the  upper  one,  we  have  6  times  5  are  30. 

Hence,  the  result  is  3025. 

The  figures  at  the  right  need  not  be  fives. 

EXAMPLE. 

64 

66 

4224  Ans. 

Here  the  rule  for  multiplying  is  the  same. 
(9) 


10  PRACTICAL  METHODS 

EXAMPLES. 


(1)  Multiply    83  by    87. 

(2)  Multiply    91  by    99. 

(3)  Multiply    28  by    22. 

(4)  Multiply  114  by  116. 


SECOND   METHOD. 

To  multiply  where  the  right-hand  figures  are  the 
same,  and  the  sum  of  the  left-hand  figures  is  ten: 

Rule. — Multiply  the  right-hand  figures  and  bring 
down  the  product.  Multiply  the  left-hand  figures  and 
add  the  figure  on  the  right  to  the  product.  Write 
this  to  the  left  of  the  former  result.  The  resulting 
number  will  be  the  required  product. 

EXAMPLE.— Multiply  46  by  66. 

46 
66 


3036  Am. 

6  times  6  are  36.  This  is  written  below.  6  times  4 
are  24,  to  which  the  right-hand  figure  6  is  added,  mak- 
ing 30.  The  result  is  3036. 


EXAMPLES. 


(5)  Multiply  34  by  74. 

(6)  Multiply  28  by  88. 

(7)  Multiply  17  by  97. 

(8)  Multiply  45  by  65. 


IN  ARITHMETIC.  \\ 

THIRD   METHOD. 

Where  numbers  are  nearly  100,  1000,  etc. 

Rule. — Add  to  the  upper  number  as  many  ciphers  as 
there  are  figures  in  the  number.  From  100  or  1000, 
etc.,  subtract  the  two  numbers  separately.  Multiply  the 
two  remainders  together,  and  place  the  product  under 
the  ciphers  at  the  right.  Then  add  the  numbers,  omit- 
ting the  final  1  at  the  left. 

EXAMPLE. — Multiply  98  by  95. 

9800 
9510 

9310  Ans. 

Here  the  two  remainders  are  2  and  5,  and  their  pro- 
duct is  10.  This  is  placed  under  the  ciphers  as  directed, 
and  the  numbers  then  added,  omitting  the  1  from  the 
19  at  the  left. 

EXAMPLE.— Multiply  994  by  982. 

994000 
982108 


97610S  Ans. 

EXAMPLES. 

(  9)   Multiply    96  by    95. 

(10)  Multiply    92  by    88. 

(11)  Multiply  995  by  990. 
Multiply  984  by  993. 


FOURTH  METHOD. 

Where  one  of  the  numbers  is  an  aliquot  part  of  100 
or  1000. 


12  PRACTICAL  METHODS 

Rule. — Multiply  by  100  or  1000,  etc.,  by  adding  two 
or  more  ciphers,  or  by  placing  the  decimal  point  two 
places  to  the  right.  Take  such  a  fractional  part  of  the 
result  as  the  multiplier  is  of  100  or  1000,  etc. 

EXAMPLE.— Multiply  336  by  25. 
4)33600 
8400  Ans. 

Since  25= J  of  100  the  product  of  336  and  25  is  J  the 
product  of  336  and  100  as  shown  in  the  operation. 

EXAMPLE.— Multiply  436  by  266|. 

Operation:  266f=2§  of  100.  Hence  436x266| 
=43600x2§=116,266§.  Ans. 

EXAMPLES. 

(13)  Multiply      72          by  125. 

(14)  Multiply  3456          by    33|. 

(15)  Multiply      75.42     by    62£. 

(16)  Multiply      31.8       by  250." 

(17)  Multiply  4072          by  833J. 

(18)  Multiply      60.732  by  444f . 


FIFTH   METHOD. 

To  square  a  number. 

Rule. — Multiply  the  number  next  larger  or  smaljer 
that  ends  in  cipher  by  one  an  equal  amount  less  or  greater 
than  the  given  number.  Add  to  the  result  the  square 
of  the  difference  between  the  given  number  and  the  one 
that  ends  in  0. 

REMARK. — For  numbers  of  two  figures  this  may  be  done  men- 
tally. 


IN  ARITHMETIC.  13 

EXAMPLE. — Square  28. 

The  numbers  to  be  multiplied  are  30  and  26.  Place 
at  the  right  4,  the  square  of  2 ;  30 — 28.  Then  3  times 
26  x  78.  Hence  the  result  is  784. 

EXAMPLE.— Square  62. 

2x2=4.     64x60+4=3844.     Ans. 

EXAMPLES. 

(19)  Square  79.  (23)   Square  103. 

(20)  Square  61.  (24)   Square    52. 

(21)  Square  48.  (25)   Square    27. 

(22)  Square  33.  (26)   Square    99. 


SIXTH   METHOD. 

To  square  a  number  ending  in  thirds. 

Rule. — Square  the  figure  on  the  left  and  add  to  the 
result  double  the  number  of  thirds  of  the  left-hand 
figure,  indicated  on  the  right.  Square  the  thirds  and 
bring  down  the  numerator  thus  obtained  twice  with  the 
same  number  of  ninths. 

EXAMPLE. — Square  63^. 

Operation:  6x6=36.  J  of  6=2.  2x2=4+36^40. 
^  of  15=!-  Hence  the  result  is  4011^.  Ans. 

EXAMPLE. — Square  36f. 

Operation:  3x3=9.  |  of  3=2.  2x2=4+9  =  13. 
|  of  |  =  |,  The  result  is  1344|.  Ans. 

EXAMPLES. 

(27)  Square  93|.  (30)   Square  126|. 

(28)  Square  96|.  (31)   Square  153,1. 

(29)  Square  331  (32)   Square    86$. 


U  PRACTICAL  METHODS 

SEVENTH   METHOD. 

To  multiply  any  numbers  by  carrying  mentally. 

Rule.  —  Multiply  the  figures  that  produce  units. 
Bring  down  into  the  column  of  products.  Multiply  all 
the  figures  that  produce  tens,  add  the  products  mentally, 
and  bring  down  the  right-hand  figure  of  the  result. 
Multiply  next  all  the  figures  that  produce  hundreds, 
add  mentally  and  bring  down  the  result,  carrying  as  in 
ordinary  multiplication.  Proceed  as  before  until  the 
multiplication  is  complete. 

EXAMPLE. — Multiply  64  by  52. 

64 
52 


3328 

Operation:     First,        2x4=8— units. 

Second,  2x6=12 
5x4=20 

20+12 =32= tens.       Write  the  2 
in  column  of  results. 

Third,     5x6+3     (carried)  =33= hun- 
dreds. 

The  total  product  is  3328.     Ans. 

EXAMPLE. — Multiply  145  by  326. 

145 

326 

47270 


IN  ARITHMETIC.  15 

Operation:     First,     5x6  =  30.     Write  the  0  in  col- 
umn of  products. 
Second,  4  x   6  =  24 
2x   5  =  10. 

24+10+  3  (carried)  =37  = 
tens.  Write  the  7  in  column  of 
products. 

Third,  Ix  6=  6. 
2x  4=  8. 
3x  5  =  15. 

15+  8+6+3    (  carried  )      =32= 
hundreds.     Write  the   2  in  column 
of  products. 
Fourth,  2  x   1   =2. 
3x   4=12. 

12+  2+  3    (carried)     =17  = 
thousands.     Write  the  7  in  column 
of  products. 
Fifth,     3x    1=   3. 

3+  1    (carried)    =4= ten 
thousands. 
The  whole  product  is  47270. 

NOTE.,— With  practice  tht,  student  may  learn  to  multiply  very 
rapidly  by  this  method,  the  work  being  entirely  mental. 

EXAMPLES. 

(33)  Multiply      45  by      82. 

(34)  Multiply      27  by      64. 

(35)  Multiply      83  by      36. 

(36)  Multiply    264  by    351. 

(37)  Multiply    746  by    825. 

(38)  Multiply  1285  by  3463. 


16  PRACTICAL  METHODS 


PRACTICAL    MEASUREMENTS. 


Nearly  all  of  the  following  rules  and  formulas  have 
been  worked  out  by  the  author  for  his  own  convenience 
in  the  class  room.  Some  of  practical  value  have  been 
taken  from  other  works  or  from  popular  custom.  They 
may  be  relied  on  as  scientifically  correct  and  accurate. 

FIRST. 
To  find  the  capacity  in  gallons  of  a  parallelopipedon. 

NOTE. — A  parallelopipeJoa  is  a  prism  whose  facjs  are  six  par- 
allelograms. A  dry  goods  box  is  an  example. 

Formula:     LxWxDx  7. 48  =  Gallons. 

Rule. — Multiply  the  length,  width,  depth  together, 
expressed  in  feet.  Multiply  this  product  by  7.48.  The 
result  will  be  the  capacity  of  the  vessel  in  gallons. 

The  product  of  length,  width  and  depth  would  give 
the  solid  contents  of  the  vessel  in  cubic  feet.  This 
multiplied  by  1728  would  give  cubic  inches,  and  this 
divided  by  231  would  give  gallons.  But  1728n-231^ 
7.48.  Hence  the  rule. 

EXAMPLE. — Find  the  capacity  in  gallons  of  a  box  8 
feet  long,  G  feet  wide,  2^  feet  deep. 

Solution:     8x6x2^x7.48=897.6  gallons.     Ans. 

EXAMPLES. 

(39)  Find  the  capacity  in  gallons  of  a  room  32  feet 
long,  20  feet  wide,  12  feet  high. 

(40)  Of  a  bin  10J  feet  long,  6|  feet  wide,  4J  feet 


IN  ARITHMETIC.  17 

SECOND. 

To  find  the  capacity  in  barrels  of  a  parallelopipedon. 
Formula:     L X  W X D X  ffi  or  ^4.2. 

Rule. — Multiply  the  product  of  the  length,  width  and 
depth,  in  feet,  by  |f,  or  divide  their  product  by  4.2. 

Since  there  are  31^  gallons  in  a  barrel  the  preceding 
formula  divided  by  31^  will  give  the  number  of  barrels. 
But  7. 48-5-31  J=fcf  Multiplying  by  £g-  is  equivalent 
to  dividing  by  4.2. 

EXAMPLES. 

(41)  Find  the  capacity  in  barrels  of  a  vessel  14  feet 
long,  12^  feet  wide,  4  feet  deep. 

(42)  Find  the  number  of  barrels  in  a  reservoir  84 
feet  long,  36  feet  wide,  18  feet  deep. 

THIRD. 

To  find  the  number  of  gallons  in  a  cylindrical  vessel. 
Formula :     Di.2  X  De.  X  V  or  5.9. 

Rule. — Square  the  diameter,  multiply  by  the  depth, 
and  this  product  by  4ff7  or  5.9. 

NOTE. — In  all  these  formulas  the  dimensions  must  be  expressed 
in  feet  before  performing  the  required  operation. 

EXPLANATION. — The  usual  way  to  obtain  this  result  is  to  find 
the  solid  contents  of  the  vessel  in  cubic  inches  and  then  divide 
by  231.  Or  the  following  formula: 

Di*X.785AxDeXl728. 
231 

But  .7864X1728=47   or  5> 
Hence  the  rule. 


18  PRACTICAL  METHODS 

EXAMPLES. 

(43)  Find  the  number  of  gallons  in  a  well  40  feet 
deep,  and  4  feet  in  diameter. 

(44)  What  is  the  capacity  in  gallons  of  a  stand  pipe 
80  feet  high,  and  10  feet  in  diameter  ? 

(45)  How  many  gallons  will  a  cistern  hold  whose 
depth  is  10  feet,  and  diameter  5  feet? 

FOURTH. 

To  find  the  number  of  barrels  in  a  cylindrical  vessel. 
Formula  :     Di.2  X  De.  X  ^  . 

47_!_Q11  —   3 

"'~O1~* 


EXAMPLES. 

(46)  Find  the  capacity  in  barrels  of  a  cistern  12 
feet  deep  and  6  feet  across. 

(47)  Find  the  number  of  barrels  in  a  cistern  14^ 
feet  deep  and  7i  feet  in  diameter,  filled  to  within  1 
foot  of  the  top. 

FIFTH. 

To  find  the  number  of  gallons  or  barrels  in  a  jug 
cistern. 

NOTE.  —  Many  cisterns  are  curved  at  the  top,  leaving  only  a 
narrow  mouth.  These  are  called  "jug  "  cisterns. 

Formula:     For  gallons,  Di.2xDe.  x5.3. 
For  barrels,    Di.2  X  De.  X  J. 

REMARK.  —  These  rules  are  only  approximate.  A  slight  variation 
from  this  may  be  made  by  the  depth  of  the  cistern,  the  curve  at 
the  top,  etc.  The  rules  given  are  derived  from  many  calculations 
of  the  actual  capacity  of  jug  cisterns.  The  mean  of  these  calcula- 
tions indicates  a  capacity  j^  less  than  cisterns  of  a  cylindrical 
shape.  The  formula  for  those  is,  Di.sxDe.x5.9;  ^  less  than  this 
would  be  Di.sxDe.  X5.3.  If  this  be  divided  by  3H,  the  quotient 
will  be  barrels;  but  5.3-*-314=£.  Hence  the  above  rule  for  barrels. 


IN  ARITHMETIC.  19 

EXAMPLES. 

(48)  Find  the  capacity  in  gallons  of  a  jug  cistern 
1 1  feet  deep  and  7  feet  in  diameter. 

(49)  A  jug  cistern  is  6|-  feet  across  and  9^  feet 
deep .     How  many  gallons  will  it  contain  ? 

How  many  barrels? 

SIXTH. 

To  find  the  quantity  of  grain  or  shelled  corn  in  a 
wagon  or  bin. 

Formula :     L  X  B  X  D  X  .8. 

Rule. — Multiply  the  continued  product  of  the  length, 
breadth  and  depth  by  .8. 

To  reduce  cubic  feet  to  bushels,  we  must  multiply 
by  1728  and  then  divide  by  2150.4,  the  number  of 
cubic  inches  in  a  bushel.  But  ^j^=.8,  nearly. 

To  find  the  number  of  bushels  of  corn  on  the  cob  the 
formula  is: 

LxBxDx.4. 

NOTE.— An  allowance  of  ^  is  made  for  the  «ob. 
EXAMPLES. 

(50)  How  many  bushels  of  grain  will  a  wagon  con- 
tain if  the  depth  is  2  feet,  the  width  3  feet,  and  length 
10  feet?     How. many  bushels  of  corn  on  the  cob? 

NOTE. — Notice  that  a  wagon  bed  10  feet  long  and  3  feet  wide 
holds  1  bushel  of  corn  on  the  cob  for  each  inch  in  height. 

(51)  How  many  bushels  of  corn  on  the  cob  in  a  bin 
16^  feet  long,  9-f  feet  wide,  6§  feet  high?    How  many 
bushels  of  grain? 

(52)  How  many  bushels  of  grain  in  an  elevator  bin 
63  feet  long,  14  feet  wide,  9  feet  high  ? 


20  PRACTICAL  METHODS 

SEVENTH. 

To  find  the  quantity  of  hay  in  a  rick. 
Formula :     I  Pis,  over + Breadth  \2  X  L, 

X  -  =  Ton, 

500 

Rule. — Find,  by  taking  the  measurement  by  a  rope, 
the  distance  over  the  stack  from  ground  to  ground. 
To  this  add  the  width.  Divide  by  4,  square  the  quo- 
tient, and  multiply  by  the  length  of  the  stack.  This 
divided  by  500  will  give  the  quantity  of  hay  in  tons. 

EXPLANATION.— By  adding  the  width  of  the  rick  to  the  distance 
over  we  get  the  entire  distance  around  the  rick.  Dividing  this  by 
4  "squares"  the  rick.  It  is  then  in  the  shape  of  a  parallelepiped, 
whose  width  and  height  are  equal,  and  whose  length  is  known. 
Multiplying  these  dimensions  together  gives  the  solid  contents  of 
the  rick  in  cubic  feet.  Dividing  this  by  the  number  of  cubic  feet 
in  a  ton  gives  the  tons  of  hay  in  the  rick. 

NOTE. — It  is  estimated  that  550  cubic  feet  of  clover,  or  450  cubic 
feet  of  timothy  are  required  for  a  ton.  The  number  here  given, 
500,  is  very  nearly  correct  for  ordinary  hay,  which  usually  consists 
of  both  timothy  and  clover. 

EXAMPLE. — Find  the  number  of  tons  of  hay  in  a  rick 
22  feet  long,  12  feet  wide,  32  feet  over. 

Solution:  32  ft.+12  ft.=44  ft.     44  ft. --4=11  ft. 
11x11x22=2662  cu.  ft. ^500^5^  tons.     Ans. 

EXAMPLES. 

(53)  Find  the  tons  of  hay  in  a  rick  30  feet  long,  14 
feet  wide,  and  30  feet  over. 

(54)  How  many  tons  of  hay  in  a  rick  13  feet  wide, 
65  feet  long,  and  27  feet  over? 

(55)  How  many  tons  in  a  rick  of  hay  42i  feet  long, 
14  feet  wide,  31|  feet  over? 


IN  ARITHMETIC.  21 

EIGHTH. 

To  find  the  quantity  o,.  hay  in  a  round  stack. 
Formula: 

(Dis.  over  -4-  A  dis.  around)2  Xi  dis.  around. 

—i-  -  2  -  ,  -==  Tons. 
4000 

Rule.  —  Add  to  the  distance  over  the  stack  from 
ground  to  ground  ^  the  distance  around  the  stack,  and 
square  the  sum.  Multiply  this  by  \  the  distance 
around  the  stack,  point  off  3  places  and  divide  by  4. 
The  result  will  be  the  quantity  of  hay  in  tons. 

EXAMPLE.  —  How  many  tons  of  hay  in  a  stack  28  feet 
over,  and  42  feet  around? 

Solution:  Distance  over  28  feet+§  of  42=42  feet. 
42X42=1764.  1764x4f  ($  of  42)=8232.  Point  off 
3  places  and  divide  by  4.  8.232-f-4=2+tons.  Ans. 

EXAMPLES. 

(56)  Find  the  number  of  tons  in  a  stack  of  hay  32 
feet  over,  and  40  feet  around. 

(57)  How  many  tons  of  hay  in  a  round  stack  26£ 
feet  over,  and  37  feet  around. 

(58)  A  round  stack  of  hay  has  the  following  dimen- 
sions: Distance  over=25  feet.     Distance  around=42 
feet.     How  many  tons  does  it  contain  ? 

NINTH. 

To  find  how  many  perch  of  rock  it  requires  to  wall 
a  well. 


Formula:     Th(Di—  Th)De_p 

~~~ 


22  PRACTICAL  METHODS 

Ride.  —  From  the  diameter  of  the  well  in  feet 
tract  the  thickness  of  the  wall,  multiply  this  by  the 
thickness  of  the  wall,  and  this  by  the  depth  of  the  well. 
The  result  divided  by  8  will  give  the  number  of  perch. 

EXAMPLE.  —  How  many  perch  would  be  required  to 
wall  a  well  33  feet  deep,  5i  feet  in  diameter,  if  the  wall 
is  1  foot  thick? 


Solution:     5  J—  1=4J.    44_189    p 


EXAMPLES. 

(59)  How  many  perch  would  be  required  to  wall  a 
well  52  feet  deep,  6^-  feet  in  diameter,  the  wall  being 
1J  feet  thick. 

(60)  A  well  is  36  feet  deep  and  4^  feet  in  diameter. 
If  a  wall  10  inches  thick  be  put  in,  how  many  perch  of 
rock  will  be  required  ? 

(61)  How  many  perch  will  wall  a  well  29  feet  deep, 
5  feet  across,  if  the  wall  is  1  foot  thick  ? 

TENTH. 

To  find  how  many  bricks  it  requires  to  wall  a  well  or 
cistern. 

Formula:     Th(Di— Th)DeX75. 

Rule. — From  the  diameter  of  the  well  or  cistern  sub- 
tract the  thickness  of  the  wall,  multiply  this  by  the 
thickness  of  the  wall,  and  the  depth  of  the  well  or  cis- 
tern. Multiply  the  last  product  by  75.  The  result 
will  be  the  number  of  bricks  required. 


IN  ARITHMETIC.  23 

EXAMPLE. — Find  the  number  of  bricks  it  will  require 
to  wall  a  cistern  whose  depth  is  12  feet,  distance  across 
CA  feet,  the  wall  to  be  8  inches  thick. 

B 

Solution:     8  in.=§  ft.     6J~§=5|. 

5|xfx  12X75=3500.     Ans. 

EXAMPLES. 

(62)  Find  how  many  bricks  it  will  require  to  wall  a 
cistern  14  feet  deep,  5|  feet  across,  the  wall  being  4 
inches  thick. 

(63)  A  well  is  18  feet  deep,  4  feet  8  inches  across. 
If  a  wall  is  put  in  4  inches  thick,  how  many  bricks 

will  it  take? 

ELEVENTH. 

To  find  the  number  of  gallons  that  would  flow 
through  a  pipe  per  minute. 

Formula :     Di2  X  V  X  -1/. 

Rule. — Square  the  diameter  of  the  tube  in  inches. 
Multiply  this  by  the  velocity  of  the  water  per  second 
in  feet,  and  this  by  *-/-  or  2.4.  The  result  will  be  the 
amount  discharged  by  the  pipe  per  minute  in  gallons. 

EXAMPLES. 

(64)  How  much  water  with  a  velocity  of  25  feet  per 
second  would  flow  through  a  pipe  2  inches  in  diameter, 
per  minute? 

(65)  How   much   water  would  be  furnished  in  12 
nours  by  an  artesian  well  4^  inches  in  diameter,  the 
water  flowing  with  a  velocity  of  12  feet  per  second? 

(66)  How    much   water   would   flow  in  24  hours 
through  a  discharge  pipe  1^  inches  across,  if  the  veloc- 
ity is  estimated  at  74-  feet  per  second? 


24 

TWELFTH. 

To  find  how  high  an  object  must  be  to  be  visible  a 
certain  distance. 

The  curvature  of  the  earth  is  8  inches  to  the  mile, 
and  varies  as  the  square  of  the  distance.  That  is  for  2 
miles  it  is  4  times  8  inches  or  32  inches.  For  3  miles 
it  is  9  X  8  inches  or  72  inches,  and  so  on  indefinitely. 
This  is  ascertained  by  a  geometrical  computation  which 
would  be  out  of  place  here.  Supposing  then  the  curva- 
ture of  the  ground  to  be  uniform  like  the  sea  level, 
with  no  hills  or  valleys,  or  other  obstructions  to  sight, 
an  object  must  be  32  inches  above  the  surface  to  be 
visible  at  the  distance  of  2  miles,  72  inches  to  be  vis- 
ible 3  miles,  etc.  From  this  we  derive  the  following 
formula: 

H=D2X§, 

in  which  H  represents  Ihe  height  of  the  object  in  feet, 
D  the  distance  in  miles,  8  inches  being  f  ft 

To  find  how  far  an  object  is  visible,  whose  height  is 
known : 

Dividing  the  above  formula  by  -|,  we  have  D3= 
Hxf.  Extracting  the  square  root,  D=/v/HXf  = 
Formula  required. 

Rule: — Extract  the  square  root  of  f  the  height  of 
the  object  in  feet.  The  result  will  be  the  distance  in 
miles  the  object  is  visible. 

EXAMPLES. 

(67)  How  high  must  an  object  be  to  be  seen  10 
miles?     12£  miles?     20  miles?     100  miles? 

(68)  How   far   is   Washington    monument   visible 
(555  feet  high)?     The  Bartholdi  Statue  of  Liberty 
(322  feet)  ?     Mt.  Chimborazo  (4  miles)  ? 


IN  ARITHMETIC.  25 

THIRTEENTH. 
To  find  how  far  a  body  will  fall  in  a  given  time : 

Formula.— D=16  t2. 

Rule. — Multiply  the  square  of  the  time  in  seconds 
by  16;  or  to  be  more  exact,  by  16.08.  The  product 
will  be  the  distance  in  feet  through  which  the  body 
will  fall. 

EXAMPLES. 

(69)  How  far  will  a  body  fall  freely  through  the 
air  in  5  seconds  ?  10  seconds  ?  30  seconds  ?  1  minute  ? 

To  find  how  long  it  will  take  a  body  to  fall  from  a 
given  height. 

Formula:     T=J  VH. 

Rule. — Take  ^  the  square  root  of  the  height  in  feet. 
The  result  will  be  the  time  in  seconds. 

EXAMPLES. — (70)  How  long  will  it  take  a  body  to 
fall  from  a  height  of  36  feet?  100  feet?  2,500  feet? 
250  feet? 

NOTE. — These  formulas  are  for  bodies  falling  in  a  vacuum. 
Practically,  the  velocity  of  a  body  falling  in  air  is  diminished  by 
friction.  Hence  the  distance  through  which  it  falls  in  a  given 
time  is  somewhat  less  than  that  given  above. 


LUMBER  MEASURE. 

In  measuring  lumber,  the  thickness  of  the  boards  is 
not  taken  into  account,  provided  it  be  one  inch  or  less. 
The  amount  of  lumber  in  a  board  is  ascertained,  there- 
fore, by  computing  the  surface  of  the  board  in  square 
feet.  The  formula  would  be 

L  X  B=No.  of  feet. 


26  PRACTICAL  METHODS 

But  when  the  stuff  is  more  than  one  inch  in  thick- 
ness, the  surface  expressed  in  feet  must  be  multiplied 
by  the  thickness  of  the  board  or  plank  in  inches. 

EXAMPLES. 

(71)  How  many  feet  of  lumber  in  16  boards,  each 
14  feet  long,  6  inches  wide,  and  1  inch  thick? 

(72)  How  many  feet  of  lumber  in  36  planks,  each 
12  feet  long,  10  inches  wide,  and  2^  inches  thick  ? 

(73)  How  many  feet  in  a  pile  of  lumber  made  up 
as  follows:  42  boards,  each  12  feet  long,  5  inches  wide, 
^  inch  thick.     24  planks,  each  10  feet  long,  12  inches 
wide,  3  inches  thick.     56  pieces,  4  inches  square  and 
9  feet  long? 

How  much  would  the  above  bill  of  lumber  cost  at 
the  following  rates: 

The  boards,  $2.25  a  hundred  feet? 

The  planks,  $1.50  a  hundred  feet? 

The  pieces,  $1.20  a  hundred  feet? 

To  find  the  largest  square  stick  that  can  be  sawed 

from  a  log. 

Rule. — Multiply  the  diameter  of  the  log  by  .  7. 
To  find  the  number  of  feet  of  lumber  in  the  largest 
square  stick  of  timber  that  can  be  sawed  from  a  log. 

Rule. — Square  the  diameter  of  the  log  in  inches  and 
divide  by  2.  The  result  will  be  the  amount  of  lumber 
in  a  12-foot  log. 

SHINGLES. 

Shingles  are  put  up  in  bunches,  each  containing  250 
shingles,  which  are  reckoned  as  4  inches  wide,  on  an 
average. 

Shingles  are  usually  laid  4^  inches  to  the  weather, 
in  which  case  the  "Carpenter's  Rule"  is  as  follows: 


IN  ARITHMETIC.  27 

900  shingles  laid  44  inches  to  the  weather  will  cover  a 
square,  or  100  square  feet. 

An  allowance  of  100  shingles  is  made  for  waste  in 
laying,  and  because  not  all  bunches  are  full. 


LATHS. 

There  are  50  laths  in  a  bunch,  each  being  4  feet 
long  and  1^  inch  wide.  They  are  usually  laid  f  of  an 
inch  apart. 

Allowing  for  the  waste  in  putting  on  the  laths,  the 
"Contractor's  Rule"  is  as  follows: 

A  bunch  of  laths  will  cover  3  square  yards. 

For  rooms  9  feet  high,  one-third  as  many  bunches 
would  be  required  as  the  distance  in  feet  around  the 

room. 

EXAMPLES. 

(74)  How  large  a  square  stick  can  be  sawed  from  a 
log  whose  diameter  is  1  foot?    16  inches?    24  inches? 

(75)  How  many  feet  of  lumber  can  be  sawed  from  a 
log  12  feet  long,  10  inches  in  diameter?    16  feet  long? 
22  feet  long? 

(76)  How  many  bunches  of  shingles  would  be  re- 
quired to  cover  a  roof,"  each  side  of  which  is  40  feet 
long,  18  feet  wide? 

To  cover  a  roof  34^  feet  long,  each  side  being  16 
feet  wide  ? 

(77)  How  many  bunches  of  laths  would  be  required 
to  cover  the  sides  and  ceiling  of  a  room  17  feet  long, 
14|  feet  wide,  9  feet  high? 

(78)  How   many  bunches  would   be   required  for 
four  rooms,  each  14  feet  long,  12  feet  4  inches  wide, 
8  feet  6  inches  high,  the  walls  only  being  lathed? 


28  PRACTICAL  ME1HODH 

COMMON    FRACTIONS. 


We  desire  to  discuss  common  fractions  only  par- 
tially. What  we  shall  say  of  them  will  be  for  older 
pupils,  as  are,  indeed,  all  the  contractions  presented 
in  this  book.  We  recommend  the  pupil  to  commit  to 
memory  the  models  given  for  the  explanation  of  mul- 
tiplication and  division  of  fractions.  The  teacher  can 
invent  a  sufficient  number  of  practical  examples  for  the 
use  of  his  classes. 


ADDITION    AND    SUBTRACTION    OF 
FRACTIONS. 

Before  fractions  can  be  added  or  subtracted  they 
must  be  reduced  to  equivalent  fractions  having  a  com- 
mon denominator.  It  is  taken  for  granted  that  the 
pupil  is  able  to  do  this.  Fractions  may  usually  be 
added  most  rapidly  in  the  following  manner: 

EXAMPLE. — Add  2^,  |,  If,  4J. 
Operation:  2A 


±L 

9A  Ans. 


'jfr 


First  add  ^,  |,  and  ^  mentally,  as  they  are  readily 
reduced  to  the  same  denominator.  The  result  is  1§. 
Then  add  §  and  f .  The  result  is  ff,  or  ly*T.  This 
makes  2  to  carry. 


IN  ARITHMETIC,  29 

EXAMPLE.—  Add  4f,  1-fr,  9T\,  7f, 
Operation  :  4  f 

IA 

9  A 


Add  f  and  f=l.     Then  A+Ty=tt- 
EXAMPLE.  —  Subtract  If  from  3-|. 

Operation:  3$ 

1  2 

2J     ^TIS. 

Subtract  without   reducing   to  improper    fractions. 


EXAMPLES. 

Find  the  sum  of: 

(79)  4|,  61,  J(  i  3$. 

(80)  8i,  5GTV  28|,  10{f 

(81)  3|,  8»,  f,  5J,  9J. 

/  (82)   ISA,  6|,  41f,  31^. 
(83)   16^,  HI,  41¥V 
0^,  8^f,  194, 


Subtract: 

(85)   2|  from  5J. 
,   (86)   3  1  from  6TV 

(87)  |  from  |. 
/(88)   13| 


30 


MULTIPLICATION    AND    DIVISION    OF 
FRACTIONS. 

PRINCIPLES: 

I.  Multiplying  the  numerator  j       MulJ.  UeB  the 

or  rt 

,.  .,.       ,,      ,  .  traction, 

dividing  the  denominator    ) 

II.       Dividing  the  numerator       j       Div-des  ^ 
multiplying  the  denominator  ) 


EXAMPLE. — Multiply  §  by  •§-. 

Model. — It  is  evident  that  if  I  multiply  §  by  4  my 
answer  will  be  f.  For  according  to  Principle  I.  multi- 
plying the  numerator  multiplies  the  value  of  the  frac- 
tion. But  I  have  not  §  to  be  multiplied  by  4,  but  by 
the  £  of  4.  Hence,  since  my  multiplier  was  5  times  too 
large,  my  answer  is  5  tiroes  too  large  and  in  order  to 
obtain  the  true  result  I  must  divide  my  present  answer 
by  5.  But  according  to  Principle  II.  multiplying  the 
denominator  divides  the  value  of  the  fraction.  There- 
fore |-h5=T8g-.  This  is  the  result  obtained  by  multi- 
plying the  numerators  and  denominators  together 
respectively. 

EXAMPLE. — Divide  f  by  f. 

Model. — It  is  evident  that  if  I  divide  |  by  2  my 
answer  will  be  T3y,  for  according  to  Principle  II.  multi- 
plying the  denominator  divides  the  value  of  the  frac- 
tion. But  I  have  not  |  to  be  divided  by  2  but  by  the 
^  of  2.  Hence,  since  my  divisor  was  3  times  too  large, 
my  answer  is  3  times  too  small,  and  in  order  to  obtain 


IN  ARITHMETIC.  31 

the  true  result  I  must  multiply  my  present  answer  by 
3.  But  according  to  Principle  I.  multiplying  the 
numerator  multiplies  the  value  of  the  fraction.  Hence, 
-j^xSrdrr^.  Ans.  This  is  equivalent  to  inverting  the 
divisor  and  multiplying,  since  we  multiplied  by  3,  the 
denominator  o£  the  divisor,  and  divided  by  2,  the 
numerator. 

When  it  is  possible  to  shorten  the  work  do  so,  either 
by  cancellation  or  by  aliquot  parts. 

EXAMPLE.  —  Multiply  f  by  §. 

Operation:  *  f  xf=ij.  Ans. 

By  cancellation. 

EXAMPLE.  —  Multiply  48  by  2|. 

Operation:  2J=J  of  10.  Hence,  48x24=480^-4 
=120.  Ans. 

EXAMPLE.  —  Divide  40  by  3|. 

Operation;  3|=*  of  10.  Hence,  40-^31=4x3 
=12.  Ans. 

Here  is  another  method  which  is  recommended  for 
most  examples  in  multiplication  of  fractions. 
EXAMPLE.  —  Multiply  4§  by  24. 
Operation:  4| 

_S 
8 

11 


_ 

11  1  Ans. 

Multiply  the  upper  number  first  by  2,  then  by 
Add  the  results.     The  product  is  as  given. 


32  PRACTICAL  METHODS 

EXAMPLES. 


Multiply  : 

W   Jby  *,',. 

(93)  OJ  by  4§. 

(90)  |,  I  TV  f 

(94)   15  ft  by  64. 

'  (91)  82,  4,  },,6§. 

r  (95)   140^  by  3T3T 

(92)  423  by  4$. 

(96)   71|byl|by3J. 

Divide  : 

(  97)   56  by  2|. 

(101)    «  by  2f. 

(  98)   13  by  3J. 

(102)   162  by  18  1. 

(  99)   138  by  5f. 

^(103)  67Abyl01TV 

•(100)   4|  by  If. 

(104)    |  by  25| 

DECIMAL    FRACTIONS. 


Students  have  opportunities  of  greatly  abridging 
their  work  in  decimal  fractions,  which  are  frequently 
long  and  tedious. 

Below  we  give  methods  in  both  multiplication  and 
division. 

EXAMPLE. 

The  compound  amount  of  $1  for  40  years,  at  6  per 
cent,  is  $10.2857179.  What  is  the  amount  for  80 
years  to  three  decimal  places  ? 

Operation:  10.2857179 

97175  8201 

10  2857 

2057 

823 

51 

7 


$105.795     Ans. 

Solution:      The  amount  will  be  the  amount  for  40 
years,  multiplied  by  itself. 


IN  ARITHMETIC.  33 

Write  the  units  figure  of  the  multiplier  under  the 
decimal  place  we  wish  to  preserve.  Write  the  other 
figures  of  the  multiplier  in  reverse  order,  as  shown  in 
the  operation.  1st.  Commence  at  the  right,  beginning 
to  multiply  with  that  figure  of  the  multiplicand  imme- 
diately above  the  one  used  as  a  multiplier.  2d.  Mul- 
tiply by  the  next  figure  of  the  multiplier,  beginning 
with  the  figure  immediately  above  it,  and  placing  the 
first  product  in  the  right  hand  column  of  results. 
Continue  until  all  the  figures  possible  have  been  used, 
placing  the  results  so  that  the  right  hand  column  may 
be  even.  Allowance  must  be  made  for  carrying  as  in 
ordinary  multiplication. 

EXAMPLES. 

Multiply  : 

(105)  10.87563  by    2.34073  to  2  places. 

(106)  16.1395    by    4.635       to  1  place. 

(107)  .06708  by  31.97        to  4  places. 

(108)  19.072      by  84.7624    to  3  places. 

(109)  The  compound  amount  of  $1,  at  8  %  for  35 
years,  is  $14.7853443.     What  is  the  amount  of  $240 
for  70  years  to  3  decimal  places. 

(110)  The  compound  amount  of  $1,  at  10  %  for  28 
years,  is  $14.4209936.     For  15  years  it  is  $4.1772482. 
How  much  would  it  be  for  43  years  to  2  decimal  places  ? 

EXAMPLE.— Divide  18.9642  by  2.607  to  three  places. 

Operation:       2.^)18.9642(7.275  Ans. 
18_249 

716 

52J. 

194 

182 


34  PRACTICAL  METHODS 

Draw  a  perpendicular  line  between  the  third  and 
fourth  decimal  places.  Divide  as  usual  till  this  line 
is  reached,  then  instead  of  bringing  down  another 
figure  or  cipher,  strike  out  one  from  the  divisor  after 
each  division  and  proceed  as  in  ordinary  division. 

The  place  where  the  line  should  be  dmwn  through 
the  dividend  is  determined  as  follows: 

Multiply  the  decimal  place  to  be  reserved  by  one  of 
the  highest  denomination  in  the  divisor.  This  will 
give  the  number  of  places  to  be  left  in  the  dividend. 

Illustration:  Divide  18.9G42  by  12.007  to  three 
places. 

.001=place  to  be  reserved. 

10=highest  denomination  in  divisor. 

10 x. 001=.01.  Hence  the  line  should  be  drawn 
through  the  dividend  after  the  second  decimal  place. 

EXAMPLES. 

Divide: 

(111)  4.562  by  8.75  to  3  places. 

(112)  20  by  3.104  to  2  places. 

(113)  .04078  by  .0036  to  1  place. 

(114)  91.875  by  15.76  to  4  places. 

(115)  3,723.84  by  96.7081  to  3  places. 

(116)  If  $1  amounts  in  twelve  years,  compound  in- 
terest at  7  per  cent.,  to  $2.2521916,  what  sum  in  the 
same   time   would   amount   to    $1,200,  reserving  two 
places  ? 


IN  ARITHMETIC.  35 

PUBLIC    SURVEYS. 


In  most  of  the  western  and  southern  states  a  system 
of  public  surveys  has  made  the  artificial  divisions  of 
land  uniform  and  definite.  This  is  of  great  import- 
ance in  keeping  a  public  record  of  land  areas  and  their 
transfer  from  one  owner  to  another,  which  is  common 
in  the  United  States.  The  method  which  has  been 
devised  makes  a  description  of  any  piece  of  land  easy 
and  expressed  in  few  words. 

Several  Principal  Meridians  are  taken  at  convenient 
intervals  from  which  to  measure  distances  East  or  West. 
The  one  from  which  lands  in  Iowa,  Missouri  and 
Minnesota  are  surveyed,  is  the  Fifth,  which  extends 
north  from  the  mouth  of  the  Arkansas  River.  For 
designating  distances  North  or  South,  a  base  line  is 
taken,  the  one  corresponding  to  the  Fifth  Meridian, 
running  West  from  the  mouth  of  the  St.  Francis  River. 

O 

The  place  where  these  two  lines  cross  is  the  starting 
point  for  the  description  of  lands  situated  in  the  above 
named  States.  The  first  row  of  townships  north  of  the 
base  line  is  called  Township  I.  North;  the  second, 
Township  II.  North,  and  so  on.  Those  south  of  the 
line  are  called  Township  I.  South ;  Township  II.  South, 
and  so  on. 

The  first  row  of  Townships  west  of  the  Principal 
Meridian  is  called  Range  I.  West ;  the  second,  Range 
II.  West,  etc.  Those  east  are  numbered  in  the  same 
way,  Range  I.  East,  Range  II.  East,  etc. 

It  is  evident  that  these  two  terms,  Range  and  Town- 
ship, locate  exactly  tlie  position  of  each  township,  and 


86  PRACTICAL  METHODS 

it  is  only  necessary  to  further  designate  by  appropriate 
names  the  precise  part  of  the  township  occupied  by  the 
land  area  we  wish  to  describe. 

Each  township  is  six  miles  square,  hence  contains  36 
square  miles.  This  however  is  not  exactly  true,  since 
the  meridians  which  are  followed  in  surveying  the 
land,  converge  as  we  go  north,  thus  making  the  north- 
ern line  of  the  township  shorter  than  the  southern. 
The  farther  north  we  go  the  greater  this  difference. 
In  Iowa  it  is  about  8  feet  to  the  mile.  If  this  differ- 
ence were  allowed  to  accumulate  the  townships  would 
continually  grow  smaller  towards  the  north.  In  order 
to  keep  them  of  uniform  size  correction  lines  are  estab- 
lished, and  the  township  corners  placed  as  upon  the 
base  line  six  miles  apart.  These  correction  lines  are 
24  miles  apirt  north  of  the  base  line,  and  30  miles 
south  of  the  base  line. 

The  township  is  divided  into  sections,  each  one  mile 
square,  which  are  numbered  from  1  to  36.  In  number- 
ing these  sections  we  begin  in  the  north-east  corner  of 
the  township,  which  is  Section  1.  Number  west  6 
sections,  then  east  6  sections,  etc.,  till  all  are  numbered. 

A  section  contains  640  acres  (with  the  exception  of 
those  that  are  fractional),  and  is  divided  into  quarters, 
N.E.,  N.  W.,  S.E.,  and  S  W.  Each  of  these,  containing 
160  acres,  is  in  turn  divided  into  halves,  E.-|,  and  "W.-J, 
by  a  line  running  north  and  south.  They  are  also 
divided  into  quarters:  N.E.  J,  N.W.  J,  S.E.  J,  and 
S.W.  £,  each  containing  40  acres.  This  method  of 
division  may  be  continued  indefinitely. 

In  describing  a  piece  of  land,  we  give  its  position  in 
the  section,  township,  then  the  number  of  township 
and  range. 


37 


LOS  ANC 

Suppose  we  wish  to  describe  the  township  marked 
with  a  star  in  Figure  1. 


COB 

REO 

TIO 

N    L/ 

INK 

6 
5 

K 

M 

* 

4a 

M 

M 

fXi 
M 

O 

M 

M 

h 

t 

3 

b 

2 

VI. 

V. 

IV. 

III. 

II. 

I. 

B  A 

S  E 

!il 

NE. 

FIGURE  1. 


The  description  would  be  as  follows:  Township  4 
North,  Kange  5  West  from  the  Fifth  Principal 
Meridian. 

TOWNSHIP. 


6 

5 

4 

3 

2 

1 

7 

8 

9 

a 
10 

11 

12 

18 

» 

17 

16 

15 

14 

13 

!19 

20 

21 

* 
22 

23 

24 

30 

29 

28 

27 

26 

25 

31 

32 

b 
33 

34 

33 

36 

FlQUBK  2. 


88  PRACTICAL  METHODS 

SECTION.  SECTION  22. 


N.  W.  '.,. 

160  A. 
N.  E.  '4. 

S.W.  ',. 

8.  E.  }i. 

= 

W 

N.  yt. 

40  A. 

* 

NN' 

b  S.  '/,. 

N.W.N.E. 

S.W 

.  Vi. 

sTw^.  s.  E. 

*' 

N.W.  h. 

N.E.k.        .^ 

+ 

S.W 

.', 

S.K,, 

* 

« 

FlOUHE  3. 


FlOUKE  4. 


In  figure  4  the  part  marked  with  a  star  would  be 
described  as  follows  : 

The  N.  E.  J  of  the  N.  E.  J  of  Section  22. 

If  we  take  the  Twp.  given  in  Fig.  1,  the  entire 
description  will  be  : 

The  N.  E.  J  of  the  N.  E.  J  of  Section  22,  Tmvp.  4 
North,  Kange  5  West  of  the  Fifth  Principal  Meridun. 

EXAMPLES. 

(117)  Describe  the  pieces  of  land  marked  in  figures 
1,  2  and  4,  by  f,  a,  b,  *. 

(118)  How  much  land  in  the  W.  J  of  the  S.  E.  J  of 
Section  20,  Township  70  North,  Range  7  West  of  the 
Fifth  Principal  Meridian? 

(119)  How  much  land  in  the  following  described 
property : 

The  N.  E.  A  of  the  N.  E.  J  of  N.  W.  ]  of  Section  17, 
Township  G7  North,  Range  ()  West? 

Also,  the  E.  £  of  the  S.  E.  J  of  Section  17,  Town- 
ship 67  North,  Range  6  West? 


IN  ARITHMETIC.  39 

There  are  many  forms  in  which  conveyances  of  Real 
Estate  may  be  expressed. 

We  give  here  a  form  common  in  Iowa: 

DEED. 

KNOW  ALL  MEN  BY  THESE  PRESENTS, 

1    That  we,  Thomas  Milton  and  Mary  Milton,  husband  and  wife, 

of  the  County  of  Davis,  and  State  of  Iowa,  for  the  consideration  of 

EIGHT  HUNDRED  DOLLARS, 

Hereby  convey  to  James  Baxter,  of  tbe  County  of  Davis,  a<  d 
State  of  Iowa,  the  following  described  Real  Estate,  situate  in  the 
County  of  Davis,  and  State  of  Iowa,  to-wit : 

The  N.  E.  ^  of  the  S.  W.  &  of  Section  7,  Township  66  Nor.h, 
Range  13  West— 40  acres. 

And  we  Warrant  the  title  of  the  same  against  all  persons  whom- 
soever. 

In  Witness  Whereof,    We  have  hereunto  set  our  hands  this 
22d  day  of  December,  1888. 

THOMAS  MILTON. 
MARY  MILTON. 
Attest : 

JOHN  SMITH,  Notary  Public. 

Appended  to  this — a  warranty  deed — is  usually 
placed  an  acknowledgment,  made  by  the  parties  sign- 
ing the  deed,  before  i  notary  public,  and  attested  by 

his  notarial  seal. 

ACKNO  WLEDGEMEN  T. 

STATE  OP  IOWA,  DAVIS  COUNTY. — SS. 

Before  me,  the  undersigned,  a  Notary  Public  in  and  for  said 
county,  personally  appeared  Thomas  Milton  and  Mary  Milton,  who 
are  personally  known  to  me  to  be  the  identical  persons  whose 
names  are  affixed  to  the  foregoing  Deed  as  grantors,  and  acknowl- 
edged the  same  to  be  their  voluntary  act  and  deed. 

Given  under  my  hand  and  official  seal  this  22d  day  of 
December,  1888. 

— • — ,)  M.  B.  HORN. 

SEAL.     >• 

(120)  Draw  up  deeds  for  the  separate  parcels  of 
land  in  the  above  examples. 


40  PRACTICAL  METHODS 


THE  METTCIC  SYSTEM. 


We  have  found  that  the  Metric  System  can  be  taught 
most  successfully  by  arranging  all  the  terms  used  into 
one  table.  There  are  only  fourteen  or  fifteen  different 
terms  used  in  this  simple  system  of  weights  and 
measures,  and  they  can  be  remembered  most  easily  by 
thus  exhibiting  them  and  their  relative  values  in  one 
group. 

The  Metric  System,  which  owes  its  great  merit  to 
its  brevity,  and  the  fact  that  it  is  a  decimal  system,  was 
originated  in  France,  about  the  time  of  the  French 
Revolution.  It  is  now  in  use  in  all  the  principal 
countries  of  Europe,  except  England,  and  was  legal- 
ized in  the  United  States  in  1867.  It  will  eventually 
become  the  standard  for  weights  and  measures  in  all 
civilized  countries,  much  to  the  simplification  of  math- 
smatical  and  commercial  operations. 

THE  METER. — An  arc  of  one  degree  on  a  meridian 
was  accurately  measured,  and  from  this  the  polar  cir- 
cumference of  the  earth  determined.  The  one  40- 
millionth  part  of  this  was  taken  as  the  standard  of  the 
new  system,  and  called  a  meter.  Its  length  is  39.37  + 
inches.  This  was  divided  into  ten,  one  hundred,  and 
one  thousand  parts,  which  were  called  respectively, 
decimeter,  centimeter  and  millimeter.  For  greater 
distances  the  meter  was  multiplied  by  10,  100,  1000 
and  10,000,  and  the  respective  results  designated  by 
the  terms  dekameter,  hectometer,  kilometer  and  myria- 
meter.  For  the  fractional  parts  Latin  prefixes  were 


IN  ARITHMETIC.  41 

taken,  ending  in  i.     For  multiples    and  Greek  prefixes, 
meaning  10,  .100,  etc.,  and  ending  in  a  or  o. 

THE  GRAM. — The  weight  of  a  cubic  centimeter  of 
distilled  water  at  its  maximum  density,  39°  Fahrenheit, 
is  the  unit  of  weight.  It  is  called  the  Gram. 

THE  LITEE. — The  Liter  is  the  measure  of  capacity. 
Its  volume  is  that  of  a  cubic  decimeter. 

THE  ARE. — The  Are  is  the  unit  for  measuring  sur- 
faces. It  is  a  square,  each  side  of  which  is  a  deka- 
meter  or  10  meters,  and  hence  contains  100  square 
meters. 

THE  STERE. — The  Ste're  is  used  for  measuring  wood. 
It  is  a  cubic  meter. 

The  Meter  is  generally  used  for  measuring  the  solid 
contents  of  a  body. 

TABLE. 

10  10          10  10  10  10        10  10  10 

Myria—  Kilo—  Hecto—  Deka— METEB.  Deci—  Centi— Milli— 
Tonneau.  Quintal.  Myria—  Kilo—  Hecto—  Deka— GRAM.    Deci—  Centi— Milli— 
Kilo—  Hecto—  D.  ka— LITER.   Deci—  Centi— Milli— 
Hecto —  ARE.  Centi— 

Deka— STERE.  Deci— 
60  8  394  5 

It  will  be  noticed  that  in  the  fourth  line  100  cen- 
tiares  are  required  to  make  1  are,  and  100  ares  to 
make  1  hectare.  In  the  others,  10  of  one  denomina- 
tion make  one  of  the  next  higher. 

Quantities  are  written  in  the  metric  system  as  in 
decimal  fractions,  and  can  be  changed  from  one  denom- 
ination to  another  by  a  change  of  reading  or  of  placing 
the  decimal  point.  Thus,  the  figures  below  the  table 
may  be  read  as'  60  hectometers,  8  dekameters,  3  meters, 
9  decimeters,  4  centimeters,  and  5  millimeters.  Or,  as 


42  PRACTICAL  METHODS 

60.83945  hectometers.     Or,  as  6083.945  meters.      Or, 
as  60839.45  decimeters,  etc. 

The  use  of  this  table  will  enable  the  pupil  to  obtain 
a  mastery  of  the  Metric  System  more  quickly  and 
easily  than  in  any  other  manner. 

REMARKS. — The  centimeter  and  millimeter  are  mostly  used  in 
measuring  very  short  distances.  The  kilometer  for  measuring 
long  distances.  The  gram  is  used  for  weighing  where  great  exact- 
ness is  required;  as  jewels,  precious  stones,  etc.  For  larger 
articles,  such  as  groceries,  the  kilogram,  or  kilo,  is  commonly  used. 
For  very  heavy  articles,  such  as  hay,  etc.,  the  tocneau,  or  metric 
ton,  is  used.  For  measuring  moderate  quantities  of  liquids  aud 
solids,  the  liter  is  most  commonly  employed.  The  hectoliter  for 
large  quantities. 

The  approximate  values  of  the  denominations  of  the 
metric,  system  most  commonly  used,  are  given  in  the 
following  table  : 

DENOMINATION.  APPROX.   VALUE. 

Meter 3  feet  3f  inches. 

Decimeter 4  inches. 

Centimeter J  inches. 

Kilometer 5  .furlongs. 

Are 4  square  rods. 

Hectare 2^  acres. 

Liter 1  quart. 

Hectoliter 2jj  bushels. 

Cubic  Meter 35|  cubic  feet. 

Stere ^  cord. 

Gram 15^  grains. 

Kilogram 2|  pounds. 

Metric  Ton 2204  pounds. 


IN  ARITHMETIC.  43 


LONGITUDE    AKD    TIME. 


The  earth  rotates  on  its  axis  once  in  24  hours. 

The  shape  of  the  earth  is  that  of  an  oblate  spheroid, 
the  polar  diameter  being  -g-J-g-  less  than  the  equatorial. 
But  since  the  places  on  the  earth's  surface  rotate  in  a 
plane  perpendicular  to  its  axis,  each  place  describes  an 
exact  circle. 

Every  circle,  however  large  or  small,  contains  SCO 
degrees. 

Hence,  every  point  on  the  earth's  surface,  except  at 
the  very  poles,  describes  every  24  hours,  a  circle  or 
360  degrees.  In  one  hour,  then,  any  point  would  pass 
through  jfa  of  360°,  or  15°.  But  1  hour=60  minutes. 
Therefore  15°,  the  space  passed  over  in  1  hour,  divided 
by  60,  will  give  the  space  passed  over  in  1  minute. 
This  is  ^  of  a  degree,  or  15 '.  This,  divided  by  60,  since 
60  seconds— 1  minute,  will  give  the  space  passed  over 
in  each  second  of  time.  The  result  is  ^',  or  15". 

From  these  facts  we  form  the  following  table: 

TABLE  FOR  LONGITUDE  AND  TIME. 

1  hour  of  time=15°  of  Longitude. 
1  min.  of  time=15'  of  Longitude. 
1  sec.  of  time=15"  of  Longitude. 

The  marks,  °,  ',  ",  indicate  degrees,  minutes  and 
seconds,  respectively,  of  longitude. 

To  determine  the  position  of  a  place  on  the  earth's 
surface,  only  two  things  are  required — its  distance  north 
or  south  from  the  equator,  and  its  distance  east  or  west 


44  PRACTICAL  METHODS 

from  a  given  meridian.  The  former  is  called  Latitude. 
The  latter,  Longitude. 

A  meridian  is  a  great  circle  passing  through  the 
poles.  It  is  evident  there  may  be  one  for  each  sepa- 
rate place  on  the  earth's  surface. 

It  is  also  evident  that  as  the  meridians  approach  the 
poles  they  converge,  since  they  all  pass  through  the 
same  point.  The  distance  between  the  meridians 
varies  as  the  cosine  of  the  latitude.  At  40°  north  or 
south  latitude  a  degree  is  only  about  f  as  long  as  at 
the  equator. 

The  length  of  a  degree  at  the  equator  is  about  69 
miles.  Hence,  at  40°  north  latitude,  it  is  about  52 
miles. 

A  point  on  the  equator  rotates  with  a  velocity  of 
1037  miles  per  hour.  At  40°  north  latitude  this 
velocity  is  only  777  miles  per  hour. 

NOTE. — If  the  earth  rotated  with  a  velocity  17  times  as  great  as 
its  present  rate  of  motion,  bodies  would  lose  their  weight.  The 
force  of  revolution  would  equal  gravity. 

Longitude  is  measured  east  and  west  180°  from 
what  is  called  the  Prime  Meridian.  The  one  usually 
taken  is  the  meridian  which  passes  through  Greenwich, 
though  the  meridians  of  Washington,  Paris,  and  Ber- 
lin are  sometimes  used  in  their  respective  countries. 

The  difference  of  longitude  between  two  places  is 
ascertained  in  various  ways.  Most  generally  it  is 
found  by  astronomical  observation  of  the  passage  of  a 
star  across  the  meridians  of  the  two  places.  The  time 
of  transit  at  each  place  is  recorded  by  means  of  the 
telegraph  at  the  other.  After  allowing  for  the  time 
necessary  for  the  transmission  of  the  electric  fluid,  the 


IN  ARITHMETIC.  4  > 

difference  of  time  between  two  places  may  be  reduced 
to  longitude  by  multiplying  by  15. 

The  earth  rotates  on  its  axis  toward  the  east. 
Hence  the  farther  east  a  place,  the  earlier  the  sun  will 
appear  to  rise,  or,  the  time  will  be  laier  than  at  a  place 
farther  west.  On  the  other  hand  the  farther  west  a 
place,  the  later  the  sun  will  appear  to  rise,  hence  the 
time  will  be  earlier  than  at  places  farther  east. 

From  the  foregoing  facts  we  deduce  the  following 
rules: 

I.     To  reduce  time  to  longitude,  multiply  by  15. 
II.     To  reduce  longitude  to  time,  divide  by  15. 

III.  To  find  the  time  at  a  place  east  of  a  given  place, 
add  their  difference  of  time. 

IV.  To  find  the  time  at  a  place  west  of  a  given  place, 
subtract  their  difference  of  time. 

The  civil  day  begins  at  midnight;  the  astronomical 
day  at  noon. 

There  must  be  some  place  where  the  day  changes 
from  one  to  the  succeeding  day.  Suppose  a  person 
start  at  noon  Wednesday,  on  a^  journey  to  the  west, 
traveling  at  the  same  rate  as  the  sun.  Wherever  he 
may  be  it  will  be  noon  to  him,  as  the  sun  will  always 
be  on  the  meridian.  In  24  hours  he  will  have  passed 
around  the  earth,  and  though  it  has  always  been  noon 
to  him,  when  he  returns  to  the  place  he  started  it  is 
not  Wednesday  noon  but  Thursday  noon.  When  and 
where  did  this  change  occur?  By  common  consent 
this  place  is  about  180°  from  Greenwich,  in  the  Pacific 
Ocean.  It  is  called  the  "International  Date  Line."  It 
is  not  a  meridian  but  an  irregular  line,  so  made  to  suit 


46  PRACTICAL  METHODS 

the  purposes  of  trade,  as  it  would  be  inconvenient  if 
the  date  line  passed  through  an  inhabited  country  or 
island. 

It  is  always  one  day  later  on  one  side  of  the  date 
line  than  on  the  other.  When  it  is  Tuesday  in 
China  it  is  Monday  in  America.  A  ship  sailing  to- 
ward China  would  skip  a  day  upon  crossing  the  line, 
one  sailing  in  the  opposite  direction  would  fall  back  a 
day. 

EXAMPLE. — The  time  at  New  York  is  9:15  A.M.  At 
Bloomfield,  la.,  it  is  8:23  A.M.  What  is  the  differ- 
ence in  longitude  between  the  two  places? 

Solution  •      9 : 15—8 : 23==52  m. 

52mxl5=13°=Dif.  in  Longitude.  Ans. 

EXAMPLE. —  The  longitude  of  St.  Petersburg  is  30° 
16 '  East.  That  of  Boston  71°  4 '  West.  What  is  their 
difference  of  time? 

Solution:     71°     4' 
30°  16' 
15)101°  2Q'=Pif.  in  Longitude. 

6hr.  45m.  20sec.— Dif.  in  Time.     Ans. 
When  places  are  on  the  same  side  of   the  prime 
meridian,  subtract  to  find  their  difference  of  longitude. 
When  they  are  on  opposite  sides  of  the  prime  meridian, 
add  to  find  their  difference  in  longitude. 

EXAMPLES. 

(121)  The    longitude  of  New  York  is  74°  24"  W. 
That  of  San  Francisco  122°  27'  49"  W.     When  it  is 
noon  at  San  Francisco,  what  time  is  it  at  New  York  ? 

(122)  The  longitude  of  Des  Moines,  la.,  is  93°  37' 
16  "  W .     That  of  St.  Petersburg  30°  16 '  E.    When  it 


IN  ARITHMETIC.  47 

is  2  o'clock  A.M.  at  St.  Petersburg,  what  time  is  it  at 
Des  Moines? 

(123)  The  difference  in  time  between  Chicago,  111., 
and  Paris,   Fr.,  is   5hr.    59in.    40  sec.     If  Chicago  is 
in    longitude  87°  35'  W.,    what  is  the    longitude  of 
Paris? 

(124)  The  longitude  of  Pittsburg,  Pa.,  is  80° 2 'W. 
If  it  is  midnight  at  St.  Louis  when  it  is  40m.  41  sec. 
A.M.  at  Pittsburg,  what  is  the  longitude  of  St.  Louis? 

(125)  The  longitude  of  Pekin,  China,  is   116°,  26' 
E.     That  of  Rio  Janeiro,  Brazil,  43°  20'  W.     When  it 
is  6  P.   M.  Tuesday  at  Pekin,  what  is  the  tin  e  at  Rio 
Janeiro  ? 

(126)  The  time  at  Washington,  longitude  77°  36" 
W.,  is  9  o'clock  A.  M.     The  time  at  Berlin  "s  1  min. 
34|  sec.  past    3    p.    M.     What    is    the    longitude  of 
Berlin? 

(127)  Bombay  is  in  longitude  72°  48 '  E.    Denver, 
Colo.,  in    longitude    105    W.     When  it  is  7  P.  M.  at 
Denver,  what  is  the  time  at  Bombay? 


RATIO. 


Ratio  is  the  numerical  relation  existing  between 
numbers  of  the  same  denomination. 

The  ratio  of  numbers  is  obtained  by  dividing  one  of 
the  numbers  by  the  other. 

The  sign  of  a  ratio  is  the  colon  (  :  ),  which  is  an 
abbreviation  of  the  sign  of  division,  the  horizontal  line 
being  omitted. 


48  PRACTICAL  METHODS 

The  antecedent  is  the  first  terra  of  the  ratio. 

The  consequent  is  the  second  term  of  the  ratio. 

Since  the  sign  of  a  ratio  is  really  a  sign  of  division, 
the  ratio  between  two  numbers  is  obtained  by  dividing 
the  antecedent  by  the  consequent. 

Thus:     4  :  6=4-=-  6,  or  §. 

A  simple  ratio  is  one  of  two  terms  ;  as,  2:3,  5:7,  etc. 

A  compound  ratio  is  the  combination  of  two  or  more 
simple  ratios. 

__  2y  3 

"*"         ' 


3  :  5 

The  ratio   can  be  found   of   numbers  of  the   same 
denomination  only. 

EXAMPLES. 
What  is  the  ratio  of  — 

(128)  1  ft.  to  1  in.? 

(129)  2  gal.  2  qt.  to  3  gal.  1  pi? 

(130)  5  hr.  24  min.  to  1  day? 

(131)  $16.50  to  $3.12|? 

(132)  2  A.  40rds.  to  20  A.? 

(133)  JtoJ? 

(134)  3T6Tto2^.? 

(135)  34§to57|? 


PROPORTION. 


An  equality  of  ratios  is  called  a  Proportion.  That 
is,  when  the  relation  existing  between  two  sets  of 
quantities  is  the  same,  the  quantities  are  said  to  be  in 
proportion. 


IN  ARITHMETIC.  49 

The  double  colon  ( : : )  is  the  sign  of  proportion. 
Thus,  4:6  : :  6:9;  which  is  read,  4  is  to  6  as  6  is  to  9, 
or  the  ratio  of  4  to  6=the  ratio  of  6  to  9. 

There  are  4  terms  to  a  proportion,  designated  in 
numerical  order:  1st,  2d,  3d,  4th.  The  first  and  fourth 
terms  are  called  the  extremes;  the  second  and  third 
the  means. 

The  great  rule  of  proportion  is  as  follows  : 

Rule. — The  product  of  the  extremes  in  a  true  pro- 
portion is  equal  to  the  product  of  the  means. 

Hence,  when  any  three  terms  are  given,  the  other 
one  may  be  easily  found. 


SIMPLE   PROPOBTION. 

Simple  Proportion  is  the  equality  of  two  simple 
ratios. 

For  finding  the  fourth  proportional  when  three  are 
given  the  following  is  the  best  method: 

Rule. — Write  for  the  Third  Term  that  number  which 
is  of  the  denomination  required  in  the  answer.  If  from 
the  nature  of  the  problem  the  answer  is  to  be  larger 
than  the  third  term,  place  the  larger  of  the  two  remain- 
ing numbers  for  the  second  term,  and  the  smaller  for 
the  first. 

If  the  answer  is  to  be  smaller  than  the  third  term 
place  the  smaller  of  the  two  numbers  for  the  second 
term,  and  the  larger  for  the  first. 

Divide  the  first  term  into  the  product  of  the  second 
and  third.  The  quotient  will  be  the  fourth  term. 


50  PRACTICAL  METHODS 

EXAMPLE. — If  $240  produces  $12  interest  in  1  year, 
Low  much  will  $400  produce? 

Solution:     240:400:  :  12:  (     ) 

Place  $12  for  the  third  term  because  it  is  of  the 
denomination  (interest)  required  in  the  answer.  If 
$240  produces  $12,  will  $400  produce  more  or  less  than 
$12?  It  will  produce  more.  Therefore  place  $400 
for  the  second  term  and  $240  for'  the  first. 

400  x  $12 

^$20.     Ans. 
240 
NOTE. — Use  cancellation  whenever  possible. 

EXAMPLES. 

(136)  A  man  earns  $2.25  by  1^  days  work.     How 
much  will  he  earn  in  10^  days? 

(137)  If  sound  travels  6,672  ft.  in  6  sec.  how  far 
will  it  travel  in  22|  sec.  ? 

(138)  A   building   42   ft.    high    casts,   at   noon,   a 
shadow  26  ft.  long.     How  high  must  a  building  be  to 
cast  a  shadow  42  ft.  long? 

(139)  If  the  velocity  of  the  electric  fluid  is  such 
that  it  would  travel  around  the  earth  (24,897  m.)  11^ 
times  in   1   sec.,  how  far  would  it  travel  at  the  same 
rate  in  235  of  a  sec.  ? 

(140)  Light  travels  from  the  sun  to  the  earth  (92,- 
000,000  m. )  in  498   sec.     How  long  would  it  bo  in 
traveling  from  the  earth  to  the  moon  (238,000  m.)  ? 

(141)  If  a  pipe  flows  118   gals,    of   water    in   6£ 
minutes,  how  many  pipes  would  it  take  to  flow  the  same 
amount  in  1    minutes? 


IN  ARITHMETIC.  51 

COMPOUND   PKOPORTION. 

Compound  proportion  is  an  equality  of  ratios,  one  or 
more  of  which  is  compound. 

Rule.  —  Place  for  the  third  term  that  number  which 
is  of  the  denomination  required  in  the  answer. 

Take  each  pair  of  numbers  and  arrange  them 
according  to  the  rule  for  simple  proportion,  always 
asking  the  question  with  regard  to  the  third  term. 

Divide  the  product  of  the  third  and  second  terms  by 
the  first  term.  The  quotient  will  be  the  fourth  term 
required. 

EXAMPLE.  —  If  4  men  in  6  days  mow  a  field  of  25 
acres,  how  long  will  it  take  7  men  to  mow  a  field  of  80 
acres  ? 


Solution:     1 


4    ::6    4x6x16 


,    .        ^xs^  =io  3  4  days.     Ara. 
•86-  16        7x5 


7x5 

Place  6  days  for  the  third  term,  because  days  is 
required  in  the  answer. 

If  it  takes  4  men  6  days  to  mow  a  field,  will  it  take 
7  men  longer,  or  not  so  long?  It  will  not  take  so 
long.  Hence,  take  4  for  the  second  term  and  7  for  the 
first. 

If  25  acres  are  mowed  in  6  days,  will  it  take  more 
or  less  than  6  days  to  mow  80  acres?  It  will  take 
longer.  Therefore  place  80  for  the  second  term  and 
25  for  the  first. 

By  multiplication  and  division,  the  result  is  10-f-f 
days. 

REMARK.— Notice  that  the  question  is  always  asked  with  regard 
to  the  3d  term,  by  each  pair  of  numbers,  without  any  reference  to 
the  other  numbers  involved. 


52  PRACTICAL  METHODS 

SOLUTION  BY  CAUSE  AND  EFFECT. 

Solutions  of  examples  in  proportion,  by  cause  and 
effect,  are  based  on  the  theory  that:  I.  Like  causes 
produce  like  effects.  II.  Causes  are  uniform  in  their 
operation.  And,  therefore:  III.  Effects  are  propor- 
tioned to  the  causes  that  produce  them. 

Rule. — Arrange  all  the  numbers  that  constitute  the 
first  cause  as  the  first  term  of  the  proportion.  Ar- 
range all  the  numbers  that  constitute  the  second  cause 
for  the  second  term.  Place  the  first  effect  for  the 
third  term,  and  the  second  effect  for  the  fourth  term. 
Divide  the  product  of  the  terms  in  which  the  unknown 
quantity  is  found  into  the  product  of  the  other  terms. 
The  result  will  be  the  required  number. 

EXAMPLE. — If  a  bin  12  ft.  long,  8  ft.  wide,  and  7  ft. 
high  contains  268  bu.  of  corn,  how  wide  must  a  bin  be 
whose  length  is  15  ft.  and  depth  6*  ft.  to  contain 
300  bu.  ? 

Solution:  The  first  cause  is  a  bin  12  ft.  long,  8  ft. 
wide,  and  7  ft.  high.  The  second  cause  is  a  bin  15  ft. 
long,  6  ft.  deep,  and  (X)  ft.  wide.  The  first  effect  is 
268  bu.  The  second  effect  300  bu.  Arrange  accord- 
ingly and  solve. 

12:  15 

8      6    ::268:300 
7    (X) 

20 

'  ===  -«<y>-  =  8f  4  ft.     Ans. 


*         X         67 

NOTE. — The  student  must  use  bis  own  judgment,  which  one  of 
the  two  methods  will  be  the  most  convenient  for  each  example. 


IN  ARITHMETIC.  53 

EXAMPLES. 

(142)  If  the  use  of  $450  for  2  yrs.  6  mo.,  at  6  per 
cent,   is  $67.50,  what  sum   would  produce  the  same 
amount  in  3  yr.  8  mo.  at  8  per  cent.  ? 

(143)  If  12  men  cut  a  pile  of  wood  96  ft.  long,  8  ft. 
high  and  4  ft.  wide,  in   2^  days,  how  long  would  it 
take  6  men  to  cut  a  pile  of  wood  38  ft.  long,  12  ft. 
wide  and  6  ft.  high? 

(144)  What  sum  of  money  in  2^  yrs.,  at  10  per 
cent.,  would  produce  3  times  as  much  interest  as  $640 
at  12  per  cent,  for  5§  yrs.  ? 

(145)  If  40  yds.  of  carpeting  are  required  to  cover 
a  floor  22  ft.  6  in.  long,  18  ft.  4  in.  wide,  how  many 
would  be  required  of  the  same  width  to  cover  a  floor 
14  ft.  6  in.  wide,  and  15  ft.  2  in.  long? 

(146)  If  42  men  dig  a  trench  365  ft.  long,  4^  ft. 
wide,  6  ft.  deep,  in  7f  days,  by  working  10J  hours  a 
day,  how  long  a  trench  3  ft.  wide  and  5  ft.  deep,  can 
18  men  dig  in  14  days,  working  9  hours  a  day? 

(147)  If  6  men  in  4|  days,  of  8  hours  each,  mow 
a  field  of  21-J  A.,  how  many  hours  a  day  would  5  men 
require,  working  9|  days,  to  mow  a  field  of  50f  A.  ? 


GENERAL  PRINCIPLES. 

As  we  stated  in  the  preface  to  this  little  volume, 
there  are  some  principles  in  Arithmetic  which  apply 
to  a  great  number  and  variety  of  subjects.  We  give 
some  of  them  below,  and  shall  constantly  use  them  in 
the  remainder  of  the  work.  The  same  conditions  con- 
stantly reappear  in  problems  under  widely  different 
subjects.  If  we  apply  the  same  rules  to  all  these  prob- 


54  PRACTICAL  METHODS 

lems  much  time  will  be  saved  from  committing  unnec- 
essary rules,  and  Arithmetic  begins  to  show  to  us  the 
unity  of  a  science,  instead  of  being  a  mass  of  details 
with  no  organic  relation  existing  between  them. 

PRINCIPLE  I. 

To  find  the  fractional  part  of  a  number: 

Multiply  by  the  fractional  part. 

This  rule  mny  be  used  in  Common  and  Decimal 
Fractious,  Percentage,  and  its  numerous  applications, 
Mensuration,  etc. 

PRINCIPLE  II. 

To  find  what  part  one  number  is  of  another: 
Divide  the  one  that's  "  is  "  by  the  one  that's  "  o/." 
Probably  no  rule  in  Arithmetic  applies  so  widely  as 
this.     It  may  be  used  in  all  the  subjects  named  above. 

PRINCIPLE  III. 

To  find  a  number  when  a  fractional  part  of  it  is 
given : 

Divide  by  the  fractional  part. 

These  rules,  in  addition  to  a  few  special  ones  belong- 
ing to  different  subjects,  are  usually  all  that  are  neces- 
sary for  one-half  the  operations  of  Arithmetic. 

They  will  be  referred  to  hereafter  as  Rules  I,  II. 
and  III. 


IN  ARITHMETIC.  55 


PERCENTAGE. 

Percentage  is  a  general  term  which  embraces  all 
calculations  in  which  100  is  taken  as  the  basis  of  com- 
parison. 

The  word  is  taken  from  two  Latin  words,  per  and 
centum,  meaning  by  the  hundred. 

The  Bate  is  the  number  of  hundredths  that  is  taken 
of  a  number.  It  is  written  as  a  decimal  fraction  of  the 
denomination  of  hundredths,  or  is  designated  by  the 
per  cent  mark,  %. 

The  Percentage  is  the  result  obtained  by  multiply- 
ing the  number  on  which  percentage  is  to  be  computed, 
by  the  rate. 

ALIQUOT  PARTS. 

Before  proceeding  to  the  discussion  of  problems,  we 
give  a  table  of  Aliquot  Parts  of  a  hundred,  by  using 
which  the  work  of  many  problems  may  be  abridged 
and  made  easier. 

TABLE  OF  ALIQUOT  PARTS. 

4  %=&.  37J%=4 

5  %-sV  40  %=£ 


10  %=rV»  50  %=i- 

iH%=f  574%=*. 

12i%=|.  60  %=£ 


20  %=\.  75  %= 

22|%=f.  Hi%= 

25  %=J. 

284  %=|. 


66 

All  the  problems  in  Percentage  may  be  solved  by  one 
or  more  of  the  general  principles  given  above.  It  is 
unnecessary  and  unwise  to  divide  the  subject  into  four 
different  cases,  as  is  common  in  most  arithmetics. 
The  mind  learns  and  remembers  with  difficulty  the 
artificial  divisions,  which  confuse  by  their  number, 
without  adding  clearness  to  the  mental  vision. 

At  any  rate,  no  more  than  three  cases  should  be 
formed,  corresponding  to  the  3  principles  given,  which 
apply  to  Percentage  and  nearly  all  its  applications. 

It  is  proposed  to  give  an  example  for  each  of  the 
four  cases  usually  treated  of,  in  order  to  show  that 
they  can  all  be  solved  by  the  above  principles. 

EXAMPLE. — Find  8  per  cent,  of  240. 

Solution:     Use  Rule  I.     240 x. 08=19. 2.     Ans. 

REMARK. — Express  the  rate  as  a  common  fraction  whenever  it  is 
a  convenient  aliquot  part  of  100. 

EXAMPLE. — Find  12i  per  cent,  of  328. 

Solution:  12^  per  cent=|-.  Rule  I.  328  X£= 
41 .  Ans. 

EXAMPLE. — Find  44|  per  cent,  of  72. 
Solution:     44^  per  cent.=£.     -|  °^  72=32.    Ans. 
EXAMPLE. — 8  is  how  many  per  cent,  of  32  ? 
Solution:     8=^  of  32.     ^=25  per  cent.     Ans. 

Or,  use  Kule  II,  The  "  is"  number  is  8.  The  "o/" 
number  32.  8-^32=25  per  cent.  Ans. 

EXAMPLE. — 18  is  how  many  per  cent,  of  156? 

Solution:  Eule  II.  18^-156=.11TV  =  11TV  per 
cent.  -4ns. 


IN  ARITHMETIC.  57 

EXAMPLE. — $3.80  is  5  per  cent,  of  what  sum? 
Solution:     Kule  III.     $3.80-f-.05=$76.     Ans. 

Or,  Rule  II.     $3.80   is    5   per  cent,   of  what  No.? 
S3. 80  -=-.05  =  $76.     Ans. 


EXAMPLE. — r2r  is  80  per  cent,  of  what  No.  ? 

Solution:  Rules  II.  or  III.  80%=f  T2T^-f=irV 
Ans. 

EXAMPLE. — f-  of  a  number  is  10.    What  is  the  No.  ? 
Solution:     Rule  III.     10-^|=16§.     Ans. 

Enough  problems  have  now  been  given  to  show  how 
the  three  rules  may  be  applied.  The  student  should 
solve  the  following  problems  in  the  same  manner. 

EXAMPLES. 

(148)  Find:     llf  %  of  $4,230. 

(149)  37^  %  of       172. 

(150)  n  %  of        §. 

(151)  IQOO"  %  of         20. 

(152)  1C6|  %  of         84. 
Solution:     166§  %=|.     f  of  84=140.     Ans. 

(153)  ,225  %  of  92.6 

(154)  10  %  of  121  %  of  22f  %  of  23.64. 

(155)  104  %  of  150  cattle. 

(156)  Find  40  %  of  22-f  %  of  15  %  of  816. 

(157)  $3.20  is  what  %  of  $2000? 

(158)  $5.12  is  what  %  of  $960? 

(159)  §  is  what  %  of  f  ? 

(160)  625  men  is  what  %  of  16,000  men? 

(161)  10  %  of  33J  %  of  75  %  of  a  number  is  what 
per  cent,  of  it  ? 


58  PRACTICAL  METHODS 

(1G2)  How   many   per   cent,   of   a    Twp.    6   miles 
square  does  a  man  own  who  has  960  A.9 

(163)  12  is  H  %  of  what  number? 

(164)  31  £  cents  is  22 1  %  of  what  sum? 

(165)  184  men  are  3^  %  of  how  many  men? 

(166)  4£  gal.  is  14  f  %  of  what? 

(167)  3|  mi.  is  87 A  %  of  what? 

(168)  $4.50  is  6  %"of  what  sum? 

(169)  14  horses   are   40    </o    of    what   number   of 
horses  ? 

SPECIAL  RULES. 

To  find  what  per  cent,  one  number  is  greater  than 
another. 

Divide  their  difference  by  the  less  number. 

EXAMPLE. — 12  is  how  many  per  cent,  greater  than 
8? 

Solution:     12—8=4.     4n-8=|  =50  %.     Ans. 

To  find  what  per   cent,    one   number   is  less  than 
another. 

Divide  their  difference  by  the  greater  number. 

EXAMPLE. — 8  is  how  many  per  cent,  less  than  12? 
Solution:     12—8=4.     4-12=i  =  33tL  %. 

EXAMPLES. 

(170)  A  receives  $2.50  for  a  day's  work,  B  receives 
$2.     How  many  per  cent,  greater  is   A's  wages  than 
B's?     How  much  less  is  B's  than  A's? 

(171)  The  ft)  Troy  contains  5,760  grains.     The  ft) 
Avoirdupois  7,000  grains.     How  many  per  cent,  greater 
is  the  ft)  Avoirdupois?     How  many  per  cent,  less  is 
thelb  Troy? 


IN  ARITHMETIC.  59 

(172)  A  has  $540,  B  has  $320.     How  many  per 
cent,  more  has  A  than  B?     How  many  per  cent,  less 
has  B  than  A? 

(173)  A  haystack  contains  5§  tons,  another  con- 
tains 3-J  tons.     What  per  cent,  larger  is  the  one  than 
the  other  ?     How  many  per  cent,  less  is  the   second 
than  the  first? 


APPLICATIONS  OF  PERCENTAGE. 

Many  subjects  in  Arithmetic  employ  the  operations 
of  Percentage  to  so  great  an  extent  as  to  be  called 
Applications  of  Percentage.  Of  these  we  shall  treat 
the  following: 

Commission  and  Brokerage. 

Profit  and  Loss. 

Stocks  and  Bonds. 

Premium  and  Discount. 

Investments.  * 

Interest.     Promissory  Notes. 

Partial  Payments. 

True  and  Bank  Discount 

Exchange. 

In  the  last  five  time  is  an  essential  element. 

In  the  first  five  it  is  not  an  essential  element. 


60  PRACTICAL  METHODS 

COMMISSION  AND  BROKERAGE- 


The  relation  of  principal  and  agent  is  quite  common 
in  the  business  world.  Indeed,  it  is  estimated  that 
the  much  larger  portion  of  business  is  carried  on  by 
means  of  agents. 

AN  AGENT  is  one  who  transacts  business  for  another. 

THE  PRINCIPAL  is  the  one  for  whom  the  agent  acts. 

Sometimes  agents  are  paid  a  certain  sum  per  day, 
month  or  year.  This  sum  is  called  salary  or  wages. 
Quite  frequently  they  receive  as  pay  a  certain  per  cent, 
of  the  proceeds  or  gross  amount  of  the  business,  which 
is  their  commission. 

A  COMMISSION  MERCHANT  is  one  who  sells  merchan- 
dise for  another. 

A  CONSIGNMENT  is  a  quantity  of  merchandise  sent  to 
a  commission  merchant  for  sale. 

THE  CONSIGNOR  is  the  person  who  sends  the  mer- 
chandise to  be  sold. 

THE  CONSIGNEE  is  the  commission  merchant,  or  the 
one  to  whom  the  goods  are  sent. 

THE  NET  PROCEEDS  is  the  sum  left  after  all  expenses 
incurred  in  selling  the  property  are  paid. 

A  BROKER  is  a  person  who  deals  in  real  estate, 
money,  stocks,  etc. 

BROKERAGE  is  the  sum  paid  to  a  broker  for  the 
transaction  of  business. 

PRINCIPLE. 

Commission  and  Brokerage  are  always  reckoned  on 
the  amount  of  sales,  or  the  amount  invested  or  col- 
lected. 


IN  ARITHMETIC.  61 

EXAMPLE. — A  lawyer  charges  10  per  cent,  for  col- 
lecting a  debt  of  $235.  What  is  his  commission? 

Solution:     Eule  I.     10   %  of  $235=$23.50.     Ans. 

EXAMPLE. — A  commission  merchant  receives  $12.40 
for  sales  of  merchandise  amounting  to  $310.  What 
per  cent,  did  he  charge  for  commission? 

Solution:  What  per  cent,  is  $12.40  of  8310?  Rule 
II.  $12. 40^ $310=.04=4.  per  cent.  Ans. 

EXAMPLE. — An  agent  received  $525  with  which  to 
purchase  goods  after  deducting  a  commission  of  5  per 
cent.  What  was  his  commission  and  how  much  did  he 
invest  ? 

Solution:  For  every  dollar  invested  in  goods  the 
agent  retained  5c. ;  hence  the  entire  cost  of  $1  of  goods 
was  $1.05.  For  $525  as  many  dollars'  worth  could  be 
purchased  as  $1.05  is  contained  in  $525=$500  amount 
invested.  Ans.  $525 — $500 =$25  commission.  Or 
5  %  of  $500 =$25  commission. 

Second  Solution:  The  cost  of  the  goods  was  100 
per  cent.,  the  commission  was  5  per  cent,  the  entire 
amount  was  105  per  cent,  or  $525.  Eule  II.  $525  is 
105  per  cent,  of  what  No.?  $525-^1.05 =$500  A.  I. 
$525 — $500 =$25  commission.  Ans. 

EXAMPLES. 

(174)  An  agent  sells  goods  amounting  to  $3,246.75, 
charging  4  per  cent,  commission.    What  did  he  receive 
and  what  were  the  net  proceeds? 

(175)  A  commission  merchant  receives  a  consign- 
ment of  $5,000  worth  of  goods.     He  sells  ^  at  a  com- 
mission  of  1^  per  cent.,  ^   at  1  per  cent.,  and   the 
remainder  at  1|  per  cent.     What  was  his  commission  ? 


02  PRACTICAL  METHODS 

(176)  A  collection  agent  charged  24-  per  cent,  for 
collecting  a  debt  of  $1,560.      What  amount  did  he 
.receive  ? 

(177)  If  $5.60  is  charged  for  collecting  a  debt  of 
$2,800,  what  is  the  rate  of  commission? 

(178)  A  commission  merchant  received  $60  for  sell- 
ing a  consignment  of  24  hhd.  (63  gal.  each)  of  wine 
at  $1.25  a  gallon.     What  was  his  rate  of  commission? 

(179)  A  principal  realized  $2,460  from  the  sale  of 
goods  amounting  to  $2,520.      What  rate    per  cent., 
commission  did  he  pay  his  agent? 

(180)  An  agent  received  $312  for  the  purchase  oi 
sugar,  after  retaining  a  commission  of  4  per  cent,  for 
buying.     At  $30  per  bbl.,  how  many  bbls.  did  he  buy? 

(181)  A  commission  merchant  remits  $840  to  his 
principal  as  the  proceeds  of  sales  after   retaining   a 
commission  of  5  per  cent.     What  was  the  amount  of 
sales  ? 

(182)  An  agent  sold  $900  worth  of  coffee,  retaining 
a  commission  of  4  per  cent,  for  selling.     He  after- 
wards invested  the  proceeds  in  sugar,  retaining  4  per 
cent,  for  buying.     How  much  was  invested  in  sugar, 
and  what  were  his  commissions  ? 

(183)  An  agent  sold  a  quantity  of  goods,  retaining 

3  per  cent,  as  his  commission.     He  then  invested  the 
proceeds,  retaining  2A  per  cent,  for  commission.     His 
entire  commission  was  $77.     What  was  the  quantity 
of  goods  sold,  and  the  net  amount  invested  ? 


IN  ARITHMETIC.  63 


PROFIT  AND  LOSS. 


All  commercial  transactions  are  made  for  the  pur- 
pose of  gain,  and  a  profit  made  by  one  party  to  a  bar- 
gain does  not  imply  a  loss  on  the  part  of  the  other. 
Most  exchanges  are  mutually  advantageous.  But  since 
fraud,  lack  of  judgment,  or  accident  may  make  losses 
possible,  and  even  with  the  most  prosperous  a  loss 
may  be  sustained  in  some  part  of  the  business,  it  is 
necessary  to  keep  a  debit  as  well  as  a  credit  side  to 
every  account. 

THE  COST  PRICE  is  what  is  paid  for  anything. 

THE  SELLING  PRICE  is  the  amount  received  for  that 
which  is  sold. 

GAIN  is  the  amount  the  selling  price  exceeds  the 
cost  price. 

Loss  is  the  amount  the  cost  price  exceeds  the  sell- 
ing price. 

Gain  and  Loss  may  be  expressed  as  so  many  per 
cent.,  or  as  so  many  dollars  in  value. 

Pupils  are  frequently  troubled  to  determine  the 
sum  on  which  gain  or  loss  is  to  be  computed.  To  aid 
them  in  this  the  following  rule  has  been  formulated: 

Gain  and  Loss  are  always  reckoned  on  the  Cost 
Price. 

PRINCIPLE. 

I.     When  you  have  given  the  C.  P.,  Multiply  \  -n 
II.     When  you  have  given  the  S.  P.,  Divide  •  }     * 


or 

1  — Loss  %. 


64  PRACTICAL  METHODS 

When  there  is  a  gain  per  cent,  this  must  be  added 
to  100  per  cent,  before  multiplying  or  dividing,  as  the 
case  may  be.  If  there  is  a  loss,  the  per  cent,  of  loss 
must  be  subtracted  from  1  before  performing  the 
required  operation. 

When  you  multiply  the  cost  price,  as  indicated 
above,  the  result  will  be  the  selling  price. 

When  you  divide  the  selling  price  according  to  the 
rule,  the  result  will  be  the  cost  price. 

If  the  cost  price  and  the  gain  or  loss  per  cent,  are 
given,  Rule  I.  applies. 

If  the  selling  price  and  the  gain  or  loss  per  cent,  are 
given,  Rule  II.  may  be  used. 

We  see  here  the  application  of  the  rules  given. 
They  are  usually  sufficient  to  solve  all  problems  in 
percentage  and  its  numerous  relations  without  the  aid 
of  any  special  principles. 

EXAMPLE. — A  man  invests  $1,360  and  gains  12%. 
What  is  his  profit? 

Solution:     Rule  I.    12  %  of  $1,360=$163.20.    Ans. 

EXAMPLE. — I  sell  an  article  which  cost  me  $4.50  so 
as  to  gain  $1.60.  What  per  cent,  did  I  make? 

Solution:  What  %  is  $1.60  of  $4.50?  Rule  II. 
$1.60-h$4.50=35f%.  Ans. 

EXAMPLE. — I  clear  $1,845  a  year  by  transactions, 
which  yield  a  profit  of  5%.  What  is  the  amount  of 
sales  ?  ' 

Solution:  $1,845  is  5%  of  what  sum?  Rule  II.  or 
III.  $l,845-=-.05=$36,900.  Ans. 


IN  ARITHMETIC.  65 

EXAMPLE. — By  selling  a  horse  for  $100 1  gain  11|  %. 
What  did  the  horse  cost  me? 

Solution:    Divide  the  S.  P.  by  1+gain  %. 
1.11|=$90.     Ans. 

Or,  SECOND  METHOD.— 100%  =  C.  P.     ll-^ 
Hence  the  S.  P.  must  have  been  111|%  of  C.  P.    $100 
is     111$%  of  what  sum?     $100-^11H%=$90.     Ans. 

EXAMPLE. — Gold  pens  are  sold  for  $3  apiece,  at  a 
loss  of  33t|  per  cent.  What  did  they  cost? 

Solution:  Divide  the  S.  P.  by  1 — loss  per  cent. 
$3-h.66§=$3-=-f=$4.50.  Ans. 

Or,  SECOND  METHOD.— 100  %  =  C.  P.  33|%=loss. 
Hence  S.  P.=66§  %  of  C.  P.  $3  is  66§  %  of  what 
sum?  Rule  II.  $3-^-66§%=$4.50.  Ans. 

EXAMPLES. 

(184)  Goods  which  cost  $2, 250  were  sold  at  a  profit 
of  14f  per  cent.     What  was  the  gain? 

(185)  A  horse  worth  $100  was  bought  for  10  per 
cent,  less  than  his  value,  and  then  sold  at  an   advance 
of  15  per  cent.      What  did  he  bring? 

(186)  Goods  which  cost  $840  were  marked  $960. 
What  was  the  rate  per  cent,  of  gain? 

(187)  A  house  and  lot  worth  $2,000  were  sold  for 
$1,850.     What  was  the  loss  per  cent? 

(188)  A  dealer  in  pork  cleared  $407.33  by  selling  a 
quantity  of  hams  at  an  advance  of  5  per  cent.,  after 
paying  expenses  to  the  amount  of  $17.67.     What  was 
the  amount  of  sales? 

(189)  A  creditor  compromised  with  a  borrower  at 
75  cents  on  the  dollar,  and  lost  thereby  $16.45.     What 
was  the  face  of  the  debt? 


60  PRACTICAL  METHODS 

(190)  Goods  marked  at  an  advance  of  16§  per  cent, 
bring  $1,400.     What  did  they  cost? 

(191)  I  sold  a  watch  for  $32  and  lost  by  the  tran- 
saction 37|  per  cent.     What  did  it  cost  me? 

(192)  If  I  exchange  100  shares  of  railroad  stock 
worth  $9,650  for  75  city  bonds  worth  120  per  cent,  on 
the  dollar,  what  per  cent,  do  I  lose  ? 

(193)  A  farmer  took  to  mill  24  bu.  of  wheat  worth 
65  cents  per  bu.     He  received   in  return  816  Ibs.  of 
flour,  the  market  value  of  which  was  $1.75  a  hundred 
Ibs.     What  per  cent,  did  he  pay  for  grinding? 


STOCKS  AND  BONDS. 


PEEMIUM  AND  DISCOUNT. 

When  a  number  of  persons  unite  for  the  transaction 
of  business  they  form  a  company. 

When  a  company  is  authorized  by  law  to  transact 
business  as  a  single  individual,  it  becomes  a  corpora- 
tion. Otherwise  it  is  simply  a  partnership,  or  joint 
stock  company. 

A  charter  is  necessary  to  the  formation  of  a  corpor- 
ation. This  is  a  written  instrument,  given  by  the 
legislature  of  the  state  in  which  the  company  is  organ- 
ized, defining  the  powers,  privileges,  and  obligations 
of  the  corporation. 

The  chief  differences  between  a  partnership  and  a 
corporation  are  as  follows: 

In  a  partnership  the  act  of  any  member  of  a  firm 
binds  the  whole  firm,  providing  it  be  in  the  line  of  the 


IN  ARITHMETIC.  67 

firm's  business.  While  the  business  of  a  corporation 
is  carried  on  wholly  by  its  officers,  which  are  regularly 
elected  by  the  stockholders. 

Again,  each  member  of  a  partnership  is  responsible 
for  the  entire  indebtedness  of  the  firm,  and  the  private 
property  of  the  partners  may  be  taken,  if  necessary,  to 
satisfy  this  indebtedness.  While,  unless  there  is  some 
special  law  in  the  state  to  the  contrary,  the  stockhold- 
ers of  a  corporation  are  liable  for  the  debts  of  the 
corporation  only  to  the  amount  of  stock  they  individu- 
ally own. 

The  amount  of  money  or  other  property  invested  by 
a  corporation  is  called  its  Capital  Stock. 

This  stock  is  usually  divided  into  shares  of  $50, 
$100,  etc. 

Those  who  own  these  shares  are  called  stockholders. 
When  a  party  acquires  stock  in  a  corporation  he  is 
given  a  Certificate  of  Stock,  showing  the  number  of 
shares  he  holds.  These  shares  or  certificates  of  stock 
can  be  transferred  from  one  person  to  another  by  sale 
or  gift. 

States,  counties,  cities,  railway  companies,  canal 
companies,  etc.,  are  some  of  the  different  kinds  of  cor- 
porations. 

Very  frequently  these  corporations  wish  to  borrow 
money.  They  do  this  by  issuing  bonds,  which  are 
sold  in  the  market  for  whatever  they  will  bring. 

A  BOND  is  the  written  or  printed  obligation  of  a  cor- 
poration, stamped  with  its  seal,  to  pay  money;  or,  in 
other  words,  it  is  the  promissory  note  of  a  corporation. 

It  is  not  necessary  that  all  the  capital  stock  of  a 
corporation  be  paid  in  at  the  commencement  of  the 


68  PRACTICAL  METHODS 

business.  Only  a  part  may  be  advanced,  the  remainder 
being  paid  by  the  stockholders  at  stated  intervals,  in 
installments.  These  installments  in  the  language  of 
corporations,  are  called  assessments. 

Assessments  may  also  be  levied  upon  the  stock- 
holders, in  proportion  to  the  amount  of  stock  owned, 
to  make  up  deficiencies  caused  by  losses  of  any  kind, 
or  to  enlarge  the  business  of  the  corporation. 

THE  NET  EARNINGS  of  a  corporation  is  the  remain- 
der left  after  deducting  the  expenses  of  the  business 
from  its  gross  receipts. 

These  are  usually  distributed  among  the  members  in 
proportion  to  the  amount  of  stock,  in  the  form  of  divi- 
dends. 

A  SURPLUS  is  usually  retained  out  of  the  net  pro- 
ceeds at  any  time,  to  discharge  the  ordinary  obligations, 
and  make  good  the  accidental  losses  of  the  corporation. 

THE  PAR  VALUE  of  bonds  is  their  face  value. 

When  a  corporation  is  prosperous,  and  is  declaring 
large  dividends,  its  stock  or  promissory  notes  (bonds) 
frequently  sell  for  more  than  their  par  or  face  value. 
They  are  then  said  to  be  at  a  Premium. 

If  the  business  of  the  corporation  is  transacted  at  a 
loss,  its  bonds  or  stock  will  probably  sell  below  par. 
When  this  is  the  case  they  are  said  to  be  at  a  Discount. 

The  market  value  of  stocks  is  what  they  sell  for  in 
the  open  market. 

Premium  is  the  amount  stocks  sell  for  above  their 
face  value. 

Discount  is  the  amount  the  face  value  of  stocks 
exceeds  their  market  value. 


IN  ARITHMETIC.  69 

PKINCIPLE. 

Premium,  Discount  and  Dividends  are  reckoned  on 
the  face  value  of  stocks. 

EXAMPLE.— The  C.  E.  I.  &  P.  Railway  Co.  have  de- 
clared a  dividend  of  7  per  cent.  What  is  my  share  if 
I  own  23  shares  of  $100  each? 

Solution:  Eule  I.  The  face  value  of  my  stock  equals 
$100x23=$2,300.  1%  of  $2,300=$161.  Ans. 

EXAMPLE, — I  own  38  shares  of  E.  R.  stock,  $50 
each.  What  rate  of  dividend  is  declared  if  my  share 

is  $76. 

Solution:  38  shares  @  $50=$1,900.  $76  is  what 
per  cent,  of  $1,900?  Rule  II.  $76^-$l,900=.04= 
4%.  Ans. 

EXAMPLE. — The  net  proceeds  of  a  bank  is  $2,340. 
If  it  declares  a  dividend  of  3  per  cent,  what  is  its  capi- 
tal stock? 

Solution:  $2,340  is  3%  of  what  sum?  Eule  II. 
$2, 340-^.  03 =$7  8,000.  Ans. 

EXAMPLE. — A  company  in  which  I  own  stock  has 
just  declared  a  dividend  of  7^  per  cent.  If  I  now  own 
$2,150  woith  of  stock,  how  much  had  I  before  the 
dividend  was  declared? 

Solution:  107-|%  of  my  former  stock = what  I  now 
own.  $2,150  is  107^  %  of  what  sum?  Eulell.  $2,150 
H-1.07J=$2,000.  Ans. 

EXAMPLES. 

(194)  The  National  Bank  of  the  Eepublic  has 
declared  a  semi-annual  dividend  of  4-|  per  cent.  If  A 


70  PRACTICAL  METHODS 

owns  31  shares  of  stock,  $50  each,  what  will  be  his 
share  of  the  gain? 

(195)  B  owns  22£  shares,  $100  each,  in  the  Conti- 
nental Insurance  Company.     If  the  company  declares 
a  dividend  of  2^  per  cent,  how  much  will  he  receive? 

(196)  I  receive  $142  as  my  share  of  dividends  on 
C.,  B.  &  Q.  B.  E.  stock.     If  I  own   35  J  shares,  $100 
each,  what  was  the  rate  declared? 

(197)  A  owns  72  shares  of  stock,  $100  each,  in  the 
Panama  Canal  Company.     If  he  receives  as  his  share 
of  dividends  $378,  what  was  the  rate? 

(198)  A  bank  declared  a  dividend  of  6f  per   eont. 
on  its  capita]  stock.     If  I  received  as  my  share  §130.'  0 
how  many  shares  of  stock,  $50  each,  did  I  own  ? 

(199)  A  person  receives  $315  as  a  3  per  cent,  divid- 
end on  stock  owned  in  a  canal  company.     How  many 
shares,  $100  each,  did  he  own  ? 

(200)  After  receiving  a  dividend  of  5  per  cent,  in 
stock   I  own   stock  to   the    amount   of   $735.       How 
much  did  I  own  before  the  dividend  was  declared  ? 

201.  A  company  in  which  I  own  stock  has  declared 
two  stock  dividends  of  3  per  cent,  and  4  per  cent.  If  I 
now  own  535f  shares,  how  many  shares  had  I  before  ? 

EXAMPLES  UNDER  PREMIUM  AND  DISCOUNT. 

EXAMPLE — I  purchased  91  shares  of  Continental 
Insurance  Company  stock,  $100  each,  at  7  per  cent, 
premium.  How  much  did  I  pay? 

Solution:  Rule  I.  91  shares  @  $100  each  =  $9100. 
100%+7%  =  107%.  107  %  of  $9100=89737.  Am. 


IN  ARITHMETIC.  71 


EXAMPLE. — Bought  28  shares  of  R.  E.  stock,  $50 
each,  at  81  per  cent,  and  sold  them  at  87^  per  cent. 
What  did  I  gain? 

Solution  :— Kule  I.  87£  °/0—  81  %=6£%=rate  of 
gain. 

6£ '%  of    28x$50=$91.     Ans. 

EXAMPLE.— I  paid  $1837.15  for  a  draft  for  $1810. 
What  was  the  rate  of  premium? 

Solution:—  Rule  II.  $1837.15— $1810=$27.15. 
What  %  is  $27.15  of  $1810? 

$27.15-h$1810=,015  =  l|  %.     Ans. 

EXAMPLE. — I  sold  stock  at  2^  per  cent,  discount, 
and  lost  $55.  How  many  shares  of  stock,  $50  each, 
did  I  have? 

Solution: — Rule  II.     $55  is  2^  %  of  what  sum  ? 

$55-r-.02J=$2200.      2200-^   50=44  shares.    Ans. 

EXAMPLE. — I  bought  Iowa  state  bonds  at  -|  per  cent, 
discount,  for  $4179.  What  was  the  face  of  the  bonds? 

Solution: — Rule  II.  I  paid  99^  %  for  the  bonds= 
$4179. 

$4179 =99|  %  of  what  sum? 

$4179-^.99|=$4200.     Ans. 

EXAMPLES. 

(202)  A  draft  for  $1500  was  sold  at  J  per  cent, 
premium.     What  was  the  sum  paid? 

(203)  R.   R.  stock  amounting  to  $3650,  and  pur- 
chased at  90^  per  cent.,  was    sold  at  78^  per  cent. 
What  was  the  loss  and  the  selling  price? 

(204)  Stock,  the  face  value  of  which  was  $2200, 
was  sold  for  $2227.50.  What  was  the  rate  of  premium? 


72  PRACTICAL  METHODS 

(205)  By  disposing  of  stock  at  an  advance  of  3i 
per  cent,  my  gain  was  $105.     What  was  the  face  of 
the  stock? 

(206)  New  York  city  bonds  at  2|  per  cent,  discount 
brought  $1945.     What  was  their  nominal  value? 

(207)  C.  &  N.-W.  K  E.  stock,  selling  in  the  mar- 
ket at  13  per  cent,  premium,  brought  $960.50.     What 
was  the  par  value  of  the  stock? 


Men  who  have  money  to  invest  usually  do  so  by  pur- 
chasing bonds  of  state,  city,  county,  etc.,  or  the  bonds 
or  stock  of  corporations. 

These  securities  are  usually  bought  and  sold  through 
an  association  of  brokers  and  dealers  in  stocks,  called 
a  stock  exchange.  The  rate  which  the  broker  charges 
varies  with  the  nature  of  the  transaction.  The  stock 
exchange  is  a  very  sensitive  register  of  the  different 
market  values  of  the  commodities  in  which  it  deals.  It 
measures  accurately  the  comparative  values  of  securi- 
ties, and  the  great  financial  operations  of  the  country. 
The  quotations  of  the  stock  exchange  are  controlled  by 
demand  and  supply,  when  not  interfered  with  by 
"corners"  and  other  artificial  restrictions,  and  in  turn 
regulate  the  market  value  of  nearly  all  stocks  in  the 
United  States.  Almost  every  important  city  in  the 
Union  has  its  stock  exchange. 

Many  investments  are  made  in  United  States  bonds. 
These  are  of  two  kinds — coupon  and  registered. 


IN  ARITHMETIC.  73 

Coupon  bonds  are  those  which  have  coupons  attached 
to  the  bottom  of  the  bond,  promising  to  pay  the  inter- 
est on  the  same  at  stated  intervals  until  the  bond  ma- 
tures. These  are  detached,  one  at  a  time,  at  the 
proper  dates,  and  presented  for  payment. 

Coupon  bonds  are  transferable. 

Begistered  bonds  are  those  where  the  name  of  the 
owner  is  recorded  in  the  office  of  the  U.  S.  Treasurer, 
and  can  be  transferred  by  indorsement. 

The  rate  of  interest  or  time  of  maturity  is  used  to 
distinguish  the  different  classes  of  bonds.  Thus  bonds 
bearing  interest  at  4  per  cent,  and  due  in  1891,  are 
spoken  of  as  U.  S.  4's  of  1891,  etc. 

The  United  States  Government  is  generally  able  to 
borrow  money  at  a  less  rate  of  interest  than  private 
individuals  or  corporations.  The  reason  is  because 
the  government  offers  the  safest  security  for  an  invest- 
ment, and  pays  its  interest  with  the  greatest  regular- 
ity. For  this  reason  also  its  bonds  command  a  pre- 
mium in  the  market,  thus  lowering  the  real  rate  of 
interest. 

The  student  should  notice  that  bonds  which  pay  the 
highest  rate  of  interest  are  not  always  the  most  pro- 
fitable investment.  Those  which  bear  a  low  rate  may 
yield  a  higher  rate  of  income  if  purchased  at  a  dis- 
count, than  those  whose  nominal  rate  of  interest  is  high. 
This  fact  will  be  fully  illustrated  in  the  problems  given 
below.  Pupils  should  study  carefully  this  subject,  as 
it  points  out  the  way  in  which  future  capital  may  be 
best  employed. 


74  PRACTICAL  METHODS 

PRINCIPLE. 

In  stock  investments,  Interest,  Premium  and  Dis- 
count are  always  reckoned  on  the  face  value. 

RULES. 

1.  When  you  have  the  amount  invested,  Divide  \  j> 

2.  When  you  have  the  face  value,  Multiply          \     ' 


remum 
or 
1  —  Discount. 

The  result  of  Eule  1  will  be  the  Face  Value. 

The  result  of  Eule  2  will  be  the  Amount  Invested. 

REMARK.  —  Sometimes  stock  is  said  to  be  purchased  at  80  per 
cent.  In  this  case  80  per  cent,  represents  1  —  Discount,  and  the 
given  Bum  should  be  at  once  multiplied  or  divided,  according  to 
the  conditions  of  the  problem. 

EXAMPLE.—  I  invest  $34000  in  C.,  B.  &  Q.  B.  E. 
stock  at  85  per  cent.  If  the  stock  yields  6  per  cent. 
per  annum,  what  will  be  my  income  ? 

Solution:  The  rule  just  given  applies:  $34000-^.85 
=  $40000  =  Face  Value. 

6  %  of  $40000=$2400.     Ans. 

Or,  by  Eule  II.  $34000  is  85  per  cent,  of  what 
sum? 

$34000--.85=:$40000. 

By  Eule  I.     6  %  of  $40000=$2400.     Ans. 

EXAMPLE.  —  Which  is  the  more  profitable,  U.  S.  4J's 
bought  at  105,  or  Panama  stock,  bearing  4  per  cent. 
interest,  and  purchased  in  the  market  at  90  per  cent.  ? 

Solution:  or    3   of  the  amount  invested  inU.  S. 


Bonds  is  returned  in  the  form  of  interest.    -g40  or  42T  of 

the  amount  invested  in  Panama  stock  is  returned  in 

the  form  of  interest.     Which  is  the  greater,  ^  or  -/$  ? 

2.    Hence  the  Panama  stock  is  the  better  investment. 


IN  ARITHMETIC.  75 

SECOND  METHOD. 

What  %  of  105  is  4£?     4J-r-105==4|  %. 
What  %  of  90  is  4?     4-^-90=4$  %. 
4 1 — 4f— J^.     Hence  the  Panama  stock  is  the  better 
investment. 

EXAMPLE. — C.  &  N.  W.  E.  E.  stock,  quoted  in  the 
market  at  79,  yields  6§  per  cent,  income.  If  A  receives 
§400  as  his  share,  what  is  the  value  of  his  stock? 

Solution:  Eules  I.  and  II.  $400  is  6§  per  cent,  of 
what  sum  ?  $400  ~  .06|  =  $6,000  ==  face  of  stock. 
$6,000  x  79  %=$4,740^value  of  stock.  Ans. 

EXAMPLE. —  If  7  per  cent.  Lake  Shore  Eailroad 
shares  are  bought  at  84  per  cent.,  what  is  the  rate  of 
income? 

Solution:  ¥7T  of  the  amount  invested  is  received  as 
interest  ¥7T  =  8|  %.  Ans. 

EXAMPLE. — How  must  I  purchase  N.  Y.  Central  E. 
E.  stock  which  yields  5  per  cent,  dividend,  so  as  to 
derive  an  income  of  8  per  cent? 

Solution:  Eule  II.  On  every  dollar,  face  value,  of 
the  stock  I  receive  5c.  By  the  conditions  of  the 
problem  this  is  to  be  8  per  cent,  of  the  investment. 
5c.  is  8  per  cent,  of  what  sum  ?  5c.  -s-8  %=.62£=  62 J 
per  cent.  Ans. 

EXAMPLES. 

(208)  I    invest      $18,400  in.  A.,  T.  &  S.  F.  E.  E. 
stock  at  92  per  cent.     If  the  stock  yields  5  per  cent, 
per  annum,  what  will  be  my  income? 

(209)  I  invest  $31,050  in  the  purchase  of  U.S.  3|'s 
at  103^  per  cent.     What  is  my  income  from  them  ? 


76  PRACTICAL  METHODS 

(210)  Denver  and  Rio  Grande   stock  yielding  7i 
per  cent,  may  be  purchased  at  98  per  cent.     Union  Pa- 
cific stock  yielding  44  per  cent,  at  76  per  cent.    Which 
should  I  prefer? 

(211)  A  receives  $360  as  his  share  of  a  5  per  cent, 
dividend  on  stock  purchased  at  107.     How  much  did 
he  have  invested? 

(212)  A  3  per  cent,  dividend  was  declared  on  stock 
which  I  purchased  at  92^  per  cent.     If  I  received  $195 
how  much  did  I  pay  for  the  stock? 

(213)  U.  S.  bonds  bearing  5  per  cent,  are  bought 
at  112^.     What  is  the  rate  of  income? 

(214)  Illinois  Central  Stock  was  purchased  at  64 
per  cent,  and  yields  2^  per  cent,  dividend.     What  is 
the  rate  of  income  ? 

(215)  What  is  the  market  value  of  stock  yielding  a 
6^  per  cent,  dividend  and  7^  per  cent,  rate  of  income  ? 

(216)  Stock  which  bears  4  per  cent,  dividend,  pro- 
duces an  income  of  5  per  cent.   What  is  its  market 
value? 


CONTRACTS. 


KEMABKS. — The  subject  of  contracts,  though  properly  belonging 
to  Commercial  Law,  is  so  intimately  connected  with  commercial 
transactions  presented  in  Arithmetic,  as  to  be  discussed  here.  We 
hardly  see  how  the  faithful  teacher  can  avoid  giving  his  Arithme- 
tic classes  the  principal  facts  in  connection  with  this  important 
branch  of  the  law. 

A  contract  is  an  agreement  between  two  or  more 
persons  to  do  or  not  to  do  some  certain  thing. 

The  following  four  things  are  requisite  for  a  valid 
contract: 


IN  ARITHMETIC.  77 

1,  Parties;  2,  consent  of  parties;  3,  consideration; 
4,  subject  matter. 

I.  PARTIES.     The  parties  to  a  contract  must  be  such 
as   can    legally  enter  into   the    agreement.     The  law 
usually  holds  that  minors,  idiots  and  lunatics  cannot 
make  a  binding  contract.     Formerly  this  list  included 
married  women,  but  at  present  they  can  hold  property 
and  make  contracts  in  nearly  all  the  States  of  the  Union. 

Minors  are  persons  under  21  years  of  age.  Their 
contracts  are  not  void,  but  voidable.  They  are  binding 
on  the  other  party  to  the  contract,  but  may  be  repu- 
diated by  the  minor.  He  can  be  held  legally  respon- 
sible, however,  for  the  following  things:  1,  clothing; 
2,  food ;  3,  lodging ;  4,  education ;  and  5,  medicine. 

II.  CONSENT  OF  PARTIES.     The  agreement  must  be 
entered  into  voluntarily.     Any  fraud,  deceit  or  force 
by  either  party  invalidates  the  contract  with  regard  to 
the  party  injured.     The  party  at  fault,  however,  can- 
not, by  disregarding  the  contract,  take  advantage  of 
his  own  wrong. 

III.  CONSIDERATION.      The    consideration    is    the 
inducement  or  reason  for  entering  into  the  contract. 
It  need  not    be    expressed    in    property.       Kelation- 
ship,    friendship,    or    gratitude    may   be   a   sufficient 
consideration  to  bind  a  contract.    Nor  is  the  amount  of 
the  consideration  essential.     Any  sum  or  thing,  pro- 
vided it  be  valuable,  may  warrant  the  execution  of  the 
contract. 

IV.  SUBJECT  MATTER  OF  CONTRACTS.     It  is  unnec- 
essary to  enter  into  a  contract  to  do  that  which  the 
law  commands.     Such  contracts  can  not  be  enforced. 
On  the  other  hand,  a  contract  to  do  something  which 


78  PRACTICAL  METHODS 

is  prohibited  by  law,  is  void.  Such  contracts  are  those 
which  are  against  public  policy,  those  which  are  for 
immoral  purposes,  or  are  fraudulent. 

A  contract  is  either  express  or  implied. 

An  express  contract  is  one  which  is  entered  into  by 
express  agreement. 

An  implied  contract  is  one  in  which  the  assent  of 
the  parties  to  the  contract,  or  some  part  of  it,  is 
presumed  from  circumstances.  As,  where  a  man  pur- 
chases a  horse,  the  law  will  presume  a  promise  to  pay 
a  reasonable  compensation,  if  a  price  has  not  been 
agreed  on.  Or,  when  one  hires  another  to  work  for 
him  and  nothing  is  said  about  wages,  there  is  an 
implied  agreement  to  pay  a  just  price  for  the  same. 


INTEREST. 


Interest  may  be  denned  as  the  remuneration  paid 
for  the  use  of  money. 

This  does  not  include  rent,  dividends,  etc.,  which 
are  the  remuneration  paid  for  the  use  of  capital  in  the 
form  of  property  other  than  money. 

INTEREST  is  common  in  all  civilized  countries. 
Formerly  the  taking  of  interest  in  any  form  was 
denominated  usury.  This  term  is  now  very  much 
restricted  in  its  meaning. 

USURY  is  interest  which  is  taken  in  excess  of  the 
rate  allowed  by  law.  A  penalty  is  attached  to  the  prac- 
tice of  usury  in  many  states.  In  others  any  rate  of 
interest  whatever  may  be  charged,  thus  doing  away 
entirely  with  usury. 


IN  ARITHMETIC.  79 

THE  LEGAL  KATE  of  interest  is  the  highest  rate 
allowed  by  law. 

THE  PRINCIPAL  is  the  sum  on  which  interest  is  to  be 
computed. 

THE  BATE  is  the  amount  which  is  charged  for  the 
use  of  $1  for  1  year. 

THE  AMOUNT  is  the  sum  of  the  principal  and  interest. 

SIMPLE  INTEREST  is  interest  on  the  principal  for  the 
given  time,  and  which  itself  cannot  draw  interest. 

COMPOUND  INTEREST  is  interest  on  the  principal,  and 
also  on  the  interest,  which,  upon  coming  due,  is  im- 
mediately converted  into  principal. 

METHODS  OF  COMPUTING  INTEREST. 

Out  of  the  many  methods  of  computing  interest,  two 
are  selected  which  are  considered  the  best.  These  are 
the  Aliquot  and  the  Twelve  per  cent,  method. 

NOTE  :  In  the  following  examples  30  days  will  be 
considered  a  month. 

EXAMPLE. — What  is  the  interest  on  $240  for  2  yrs. 
4  m.,  15  da.,  at  0  per  cent? 

Solution:     By  aliquot  parts. 

6  %  of  $240  =  $14. 40  =  Interest  for  1  yr. 


*=J 


28.80=         "       "    2    " 
4.80=         "       "    4  m. 
.60=         "       "15  da. 


$34.20.     Ans. 

EXAMPLE. — Find  the  interest  at  8  per  cent,  on  $165 
for  9  m.  27  da. 


80  PRACTICAL  METHODS 

Solution:  Twelve  per  cent,  method.  The  interest 
on  $1  at  12  %  for  9  m.=.09.  For  '27  da.=.009.  Hence 
for  9  m.  27  da.=.099.  8  =  f  of  12.  Int.  at  8  %  =  § 
of  .099=.066.  Int.  on  $165  =  165  x. 066--=  $10.89. 
Ans. 

EXPLANATION  OF  12  PER  CENT.  RULE. 

12  %  per  annum  =  1  %  per  m.,  since  1  yr.=  12  m. 
1%  for  1  m.  or  30  da.=.OOOJ  for  1  da.  Hence  the 
rule. 

RULE. — The  number  of  months  expressed  in  cents, 
and  ^  the  number  of  days  expressed  in  mills  equals  the 
interest  on  $1  at  12%  for  the  given  time. 

EXAMPLE. — Find  the  simple  interest  on  $1348.25  for 
1  yr.,  7  m.,  18  da.,  at  12%. 

Solution:  1  yr.  7  m.=19  m.  Interest  for  19  m.= 
.19.  Interest  for  18  da. =.006.  Hence  interest  for  1 
yr.  7  m.  18  da.  at  12  %  on  $1=.196.  On  $1348.25 
it  would  be  1348.25  X.196=$264.257.  Ans. 

EXAMPLE. — How  long  will  it  take  $640  to  produce 
$120,  simple  interest,  at  6%  ? 

Solution:  Interest  on  $640  for  1  yr.  at  6%  =$38.40. 
$120-$38.40-=3i  yr.=3  yr.  1  m.  15  da.  Ans. 

EXAMPLE. — What  is  the  rate  of  interest  when  $1665 
produces  $116.55  annually. 

Solution:  Rule  II.     $116.55  is  what  %  of  $1665? 
116.55--   1665=.07=7%.     Ans. 

EXAMPLE. — What  principal  will  produce  $298.35  a 
year  at  9%? 


IN  ARITHMETIC.  81 

Solution:    Bule  II.    1298.35  is  9%   of  what  sum? 
.35-r-.09=$3315.     -4ns. 


EXAMPLE. — What  principal  will  amount  in  2  yr.  1  m. 
6  da.  at  1%  to  $1376.40? 

Solution:  Eule  II.  Amount  of  $1  at  7%  for  2  yr.  1 
m.  6  da.=1.147.  $1376.40=114.7%  of  what  sum? 
$1376.40^-114.7%=:$1200.  Ans. 

EXAMPLES. 

NOTE. — To  get  exact  or  accurate  interest  365  days  should  be 
counted  a  year.  The  interest  on  any  sum  counting  this  number  of 
days  will  be  -7V  less  than  by  using  360  days.  To  get  exact  interest, 
therefore,  it  is  only  necessary  to  diminish  the  interest  obtained  in 
the  usual  way  by  -fa  of  itself. 

(217)  Find  the  simple  interest  on  $298.77   for  2| 
yrs.  at  6  %. 

(218)  $112.60  for  4  yr.  7  m.  14  da.  at  8  %. 

(219)  $2,356  for  9  m.  27  da.  at  8  %. 

(220)  $1,749.32  for  1  yr.  2  m.  25  da.  at  9  %. 

(221)  $750.70  for  3  yr.  11  m.  29  da.  at  10  %. 

(222)  How  long  will  it  take  $500  to  produce  $50 
at  8  %  ? 

(223)  $244  to  produce  $32  at  6  %  ? 

(224)  $974.60  to  produce  $118.20  at  8  %  ? 

(225)  What  is  the  rate  when  $400  produces  $28  in 
lyr.? 

(226)  $620  produces  $28.93  in  8  mo.? 

(227)  $2,350  produces  $302.36  in  1  yr.  7  m.  9  da.? 

(228)  What  principal  will  produce  $225.63  in  1  yr. 
at  9  %  ? 

(229)  $29.232  in  5  m.  24  da.  at  9  %  ? 

(230)  $139.725  in  2  yr.  4  m.  18  da.  at  8  %? 


82  PRACTICAL  METHODS 

(231)  What  principal  will  amount  to  $059.10  in  1 
yr.  7  m.  21  da.  at  6  %  ? 

(232)  To  $512.85  in  10  m.  10  da.  at  8  %  ? 

(233)  To  $566.57  in  4  yr.  1  m.  24  da.  at  9  %  ? 

(234)  To  $2614.21  in  3  yr.  15  da.  at  6  %  ? 

(235)  Find  the  exact  interest  on  $519.20  at  7  % 
for  2  yr. 

(236)  $123.80  at  8  %  for  9  m.  3  da. 

(237)  $4075  at  8  %  for  3  m. 


PROMISSORY  NOTES. 


A  PBOMISSORY  NOTE  is  a  written  promise  to  pay 
money  at  some  future  time. 

THE  MAKER  of  a  note  is  the  party  who  makes  him- 
self responsible  for  its  payment  by  signing  it. 

THE  PAYEE  is  the  party  to  whom  the  promise  of 
payment  is  made. 

The  one  who  owns  the  note  at  any  time  is  the  Holder. 

AN  INDOBSER  is  one  who  places  his  signature  on  the 
back  of  the  note. 

INDORSEMENT  is  for  the  purpose  of  transferring  the 
note  from  one  person  to  another. 

REMARK.— If  the  name  alone  is  written,  this  is  a  blank  indorse- 
ment, and  holds  the  indorser  responsible  for  the  payment  of  the 
note  in  case  the  maker  fails  to  pay  it. 

If  the  indorser  wishes  to  free  himself  from  such 
liability,  he  may  write  after  his  name  the  words,  "  with- 
out recourse." 

Another  method  of  indorsement  is,  pay  to  "  John  J. 
Brown."  This  is  a  "special"  indorsement. 


IN  ARITHMETIC.  83 

Promissory  Notes  are  either  Negotiable  or  Non- 
negotiable. 

A  negotiable  note  is  one  that  can  be  transferred  from 
one  person  to  another. 

A  non-negotiable  note  is  one  which  can  be  collected 
only  by  the  payee. 

The  words  "or  order"  or  "or bearer"  are  necessary 
to  render  a  note  negotiable. 

When  a  note  is  drawn  up  to  bearer,  any  person  into 
whose  possession  the  note  comes  legally  can  collect  it, 
and  no  indorsement  is  necessary.  If  it  is  drawn  up  to 
order  it  can  be  transferred  only  by  the  indorsement  of 
the  party  io  whom  it  is  made  payable,  or  the  holder. 

If  a  note  is  not  paid  when  due,  it  is  protested — -that 
is,  a  written  notice  of  the  fact  of  non-payment  is  made 
out  by  a  notary  public  and  sent  to  the  indorsers  and 
security.  If  this  is  done  they  become  liable  for  its 
payment.  If  not,  they  are  relieved  from  all  liability 
for  its  discharge. 

The  Maturity  of  a  note  is  the  date  on  which  it  falls 
due.  It  is  nominally  due  on  the  expiration  of  the  time 
mentioned  in  the  note.  It  is  not  legally  due  for  three 
days  thereafter.  These  three  days  are  called  "days  of 
grace." 

To  find  the  time  when  a  note  matures:  If  the  time 
to  elapse  before  payment  is  given  in  years  and  months, 
add  the  number  of  years  and  calendar  months  to  the 
date  of  the  note.  If  the  time  is  given  in  days,  the 
actual  number  of  days  must  be  counted  from  the  date 
of  the  note,  to  which  three  days  are  added  for  grace. 
For  example :  If  a  note  is  dated  Jan.  3d  to  run  two 
months,  it  will  be  due  March  3/e.  If  it  is  to  ran  60 


84  PRACTICAL  METHODS 

days  it  will  be  due  60  days  after  Jan .  3,  or  March 

It  is  not  necessary  to  insert  the  words,  "  for  value 
received,"  to  make  a  note  valid. 

The  amount  in  the  body  of  the  note  should  be 
written  in  words,  to  prevent  alteration  or  fraud  of  any 
kind. 

If  the  words  "  with  interest "  are  found  in  a  note,  it 
draws  interest  from  the  date  of  the  note,  and  a  rate 
fixed  by  law  is  allowed  if  no  rate  be  mentioned. 

If  the  note  does  not  contain  the  words  "with  inter- 
est," it  draws  interest  only  after  maturity. 

In  many  promissory  notes  the  rate  of  interest  is  not 
specified.  In  all  states  a  rate  is  fixed  by  law,  which 
shall  be  charged  in  such  cases  upon  the  maturity  of 
the  note,  or  from  the  time  specified  therein. 

The    following   are   the    common   forms   used    for 

promissory  notes: 

I. 
8317.20.  BLOOMFIELD,  IA.,  Dec.  27, 1888. 

Two  months  after  date,  I  promise  to  pay  to  the  order  of  H.  E. 
Hibbard,  three  hundred  and  seventeen  and  fifo  dollars,  with  inter- 
est at  6  per  cent.  B.  A.  GO  AN. 

II. 

$250.  DBS  MOINBS,  IA.,  Aug.  1, 1885. 

On  demand,  I  promise  to  pay  to  J.  M.  Howie,  or  bearer,  two 
hundred  and  fifty  and  A%  dollars. 

B.  M.  WORTHINGTON. 

III. 

865.00.  CHICAGO,  ILL.,  May  2, 1880. 

Sixty  days  after  date,  we,  or  either  of  us,  promise  to  pay  Jno.  S. 
Woolson,  or  bearer,  sixty-five  dollars,  for  value  received,  with 
interest.  FRANK  NELSON. 

G.  M.  SULLIVAN. 

The  above  is  a  joint  note,  and  the  makers  are  jointly 

liable  for  its  payment 


IN  ARITHMETIC.  85 

EXAMPLES. 

Find  the  date  of  maturity  and  the  amount  of  each  of 
the  following  notes : 

(238)  $190.60.  GALESBCBG,  IM,.,  July  4, 1879. 
For  value  received  I  promise  to  pay,  two  months  after  date,  to 

W.  H.  Sadler,  one  hundred  and  ninety  and  ^   dollars,  with 
interest  at  9  per  cent.  J.  R.  WILCOX. 

(239)  $624.00.  MADISON,  Wis.,  Oct.  18, 1887. 
On  or  before  the  tenth  day  of  Feb.,  1889, 1  promise  to  pay  U. 

S.  Miller,  or  order,  six  hundred  and  twenty-four  dollars,  with  in- 
terest at  8  per  cent.  JOHN  MARTIN. 

(240)  $2675.80.  ST.  PATO,  MINN.,  Feb.  10, 1884. 
Ninety  days  after  date  I  promise  to  pay  Chas.  Claghorn,  or 

bearer,  two  thousand  six  hundred  and  seventy-five  and  fflg  dollars, 
value  received,  with  interest  at  10  per  cent. 

C.  C.  COCHEAN. 


PARTIAL  PAYMENTS. 


When  payments  are  made  at  various  intervals  on 
promissory  notes,  they  are  called  Partial  Payments. 

There  are  various  methods  of  computing  the  amount 
due  on  a  note  when  partial  payments  have  been  made. 

The  only  two  which  we  care  to  mention  here  are  the 
United  States  and  the  Mercantile  rules. 

UNITED  STATES  RULE. 

Compute  the  interest  on  the  principal  from  the  date 
when  the  note  begins  to  bear  interest  till  the  time  of 
the  first  payment,  add  the  interest  to  the  principal  and 
subtract  the  payment. 

Compute  the  interest  on  this  sum  as  a  principal  until 
the  next  payment.  Add  the  interest  and  subtract  the 


86  PRACTICAL  METHODS 

payment  as  before.     Continue  in  this  manner  till  the 
date  of  settlement. 

Provided,  that  no  interest  shall  be  added  which  is  in 
excess  of  the  payment  made  at  that  time.  In  case  the 
interest  at  any  time  exceeds  the  payment,  interest  is  to 
be  computed  on  the  original  sum  to  such  a  time  that 
the  sum  of  the  payments  may  exceed  the  interest. 

It  is  the  policy  of  the  law  to  prevent  the  payment  of 
compound  interest,  unless  a  special  contract  is  made 
that  unpaid  interest  shall  become  a  part  of  the  principal. 
This,  of  course,  is  to  the  debtor's  advantage. 

For  example:  If  a  man  borrows  $100  at  6  per  cent, 
for  five  years,  by  simple  interest  he  should  pay  just  $6 
a  year  or  $30  at  the  end  of  the  five  years.  It  is  evi- 
dent that  if,  at  any  time,  any  of  the  interest  be  added 
to  the  principal,  more  than  $6  of  interest  will  accrue 
in  a  year,  and  to  this  extent  will  be  more  than 
simple  interest.  Now,  this  is  the  very  thing  done  by 
the  United  States  rule.  Every  time  a  payment  is 
made  the  interest  which  has  accrued  at  that  time  is 
added  to  the  principal  before  the  payment  is  deducted. 
The  oftener  payments  are  made  the  oftener  is  interest 
added  to  the  former  principal,  thus  making  a  rapid  ac- 
cumulation of  the  debt  and  putting  an  extra  burden 
upon  the  debtor  for  his  promptness.  If  payments  are 
made  at  short  intervals  compound  interest  in  its  most 
aggravated  form  will  be  the  result  of  applying  the  U.S. 
rule. 

EXAMPLE. — 
$640.  OMAHA,  NEB.,  Aug.  12, 1883. 

Three  months  after  date  I  promise  to  pay  to  John  W.  Graham 
or  order,  six  hundred  and  forty  dollars,  with  interest  at  6  per  cent 

ROBERT  J.  BROWN. 


IN  ARITHMETIC. 


87 


INDORSEMENTS. 

Dec.  12,  1883,  $25;  May  30,  1884,  $110;  Sept.  3, 
1884,  $200.     What  was  due  Jan.  1,  1885  ? 

Solution  : 

Time  from  Aug.  12  to  Dec.  12, 1883=4  mo. 

Int.  on  $  1  for  4  m.  at  12  per  ct.=.04.  For  6  per  ct.=.02. 

Int.= 


Payment= 

Time  from  Dec.  12, 1883,  to  May  30, 1884=5  m.  18  da. 

Int.  at  12  per  ct.  for  5  m.  18  da.=.056.  Int.= 

At  6  per  ct.=.028. 

Payment= 

Time  from  May  30, 1884,  to  Sept.  3, 1884=3  m.  3  da. 
Int.  at  6  per  cent,  for  3  m.  3  da.=.0155. 

Int.= 


6.40 
.02 

12.80 
640. 

652.80 
25. 

627.80 
.028 

17.578 
627.80 

645.378 
110. 

535.378 
.0155 

8.297 
535.378 


543.675 
Payment^    200. 

Time  from  Sept.  3, 1884,  to  Jan.  1, 1885=3  m.  28  da.         343.675 
Int.  at  6  per  cent.=.019f.  .019$ 

Int,=          6.759 
343.675 


Amount  due  Jan.  1, 1885. 


$350.43 


Ans. 


EXAMPLES. 

( 241 )     $2205.  ST.  Louis,  Mo.,  July  5, 1880. 

Two  years  after  date  I  promipe  to  pay  to  the  order  of  James 
Stewart,  the  sum  of  two  thousand  two  hundred  and  five  dollars, 
with  interest  at  8  per  cent.  CHARLES  MILBURN. 


INDORSEMENTS. 

Dec.  16,  1880,  $250;    May  1,  1881, 
1881,  $65;  Feb.  28,  1882, 
What  is  due  July  7,  1882? 


.30;  July  1, 


88  PRACTICAL  METHODS 

( 242)      $134.40.  BLOOMFIELD,  IA.,  Jan.  2, 1883. 

Thirty  days  after  date  I  promise  to  pay  to  Madison  Turner,  or 
order,  one  hundred  and  thirty-four  and  ^  dollars,  value  received, 
with  interest  at  6  per  cent.  THOMAS  TAYLOR. 

INDORSEMENTS. 

Jan.  20,  1883,  $5;  Feb.  10,  1883,  $40;  March  31, 
1883,  $81. 

What  was  due  April  2,  1883. 


COMMEKCIAL   EULE. 

Compute  the  interest  on  the  principal  till  the  date 
of  settlement.  Add  the  interest  to  the  principal. 

Find  the  interest  on  each  payment  from  the  time  it 
was  made  to  the  date  of  settlement.  Find  the  amount 
of  all  the  payments  and  subtract  from  the  amount  of 
the  principal.  The  balance  will  be  the  sum  due. 

NOTE. — In  computing  interest  by  this  rule  the  exact  number  of 
days  should  be  used. 

The  Commercial  rule  is  based  on  the  following  prin- 
ciple: If  a  man  borrows  a  sum  of  money  he  should 
pay  interest  on  it  until  it  is  paid,  or  until  the  date  of 
settlement.  If  payments  are  made  the  creditor  should 
pay  interest  to  the  debtor  on  these  sums  from  the  time 
they  are  paid  till  the  date  of  settlement. 

This  rule  prevents  any  compounding  of  interest,  and 
is  absolutely  fair  to  both  debtor  and  creditor,  as  it 
makes  each  party  pay  interest  on  the  money  received 
from  the  other  from  the  time  of  its  receipt  till  the  date 
of  settlement 

The  Mercantile  rule  is  generally  used  in  business  for 
periods  less  than  one  year. 


IN  ARITHMETIC.  89 

EXAMPLES. 

(243)  $850.  MEMPHIS,  TENN.,  Oct.  8, 1887. 
One  year  after  date  I  promise  to  pay  to  Sidney  Smith  or  order, 

eight  hundred  and  fifty  dollars,  value  received,  with  interest  at 
9  per  cent.  DOUGLAS  JERROLD. 

INDORSEMENTS. 

Jan  15,  1888,  $116;    Jan.   30,  1888,  $62.50;   Sept. 
1,  1888,  $180. 

Find  the  amount  due  Oct.  1,  1888. 

(244)  Find   the    amount  due  on  each  of  the  ex- 
amples in  the  previous  article,  by  the  Mercantile  rule. 


TRUE  DISCOUNT. 


Notes  are  frequently  transferred  from  one  party  to 
another  before  maturity.  The  business  of  cashing 
notes  is  one  of  the  sources  of  revenue  to  banks,  and 
money  lenders  in  general.  It  is  evident,  that  if  cash 
is  paid  for  a  note  before  it  is  due,  a  sum  should  be 
deducted  for  advancing  the  money.  The  sum  deducted 
is  called  discount. 

Since  the  party  selling  the  note  has  the  use  of  the 
money  paid  from  the  time  of  payment  till  the  time  the 
note  is  due,  the  amount  he  deducts  should  be  the  in- 
terest for  this  period  on  the  sum  received.  Or  in  other 
words,  the  sum  he  receives  for  the  note  should  be 
such  that  if  put  on  interest  it  would  amount  to  the  face 
of  the  note  at  the  date  of  its  maturity.  Or,  from  the 
standpoint  of  the  creditor,  how  much  could  he  afford 
to  pay  at  the  present  time  for  a  note  due  one  year 
hence?  Just  the  sum  which,  put  on  interest,  would 


90 

amount  in  one  year  to  the  amount  he  will  receive  for 
the  note  when  it  is  paid  at  the  end  of  the  year. 

This  sum  is  called  the  Present  Worth. 

True  Discount  is  the  difference  between  the  present 
worth  and  the  face  or  amount  of  the  debt. 

RULE  FOE  TRUE  DISCOUNT. 

Divide  the  face  or  amount  of  the  debt  by  the  amount 
of  $1  for  the  given  time  and  rate.  The  result  will  be 
the  present  worth.  This  subtracted  from  the  face  or 
amount  of  the  debt  will  give  the  true  discount. 

Ex  AMPLE.- -A  note  for  $325,  due  in  3  mo.,  was  sold 
for  cash.  If  money  was  worth  8  per  cent.,  how  much 
was  paid  and  what  was  the  true  discount  ? 

Solution:  $1  in  3  mo.  at  8  %  would  amount  to 
$1.020§.  It  would  take  as  many  dollars  to  amount  to 
$325  as  1.020§  is  contained  in  $325  =  $318.42  P.  W. 
Ans.  $325— $318. 42  =  $6.58  Dis.  Ans. 

EXAMPLE. — A  note  due  in  1  yr.  4  m.  and  drawing 
interest  at  9  per  cent,  was  discounted  at  8  per  cent. 
What  was  the  amount  paid  ? 

Solution:  Amount  of  $260  for  1  yr.  4  mo.  and  3  da. 
at  9  %  =  $291.395.  This  is  the  amount  to  be  re- 
ceived at  the  maturity  of  the  note,  hence  is  the  sum  to 
be  discounted. 

Ami  of  $1  for  1  yr.  4  mo.  8  da.  at  8  %  =$1.107j. 
$291. 395^-1.107i  =  $263. 15.  Ans. 

PROBLEMS. 

(245)  A  note  for  $450,  running  2  yr.  3  mo.  16  da., 
was  discounted  at  7  per  cent.  What  was  the  amount 
paid? 


IN  ARITHMETIC.  91 

(246)  A  note  for  $1, 264  80,  bearing  6  per  cent, 
interest,  dated  Jan.  1,  1885,  and  due  Jan.  1,  1880,  was 
sold  Aug.  10,  1885,  discount  off  at  8  per  cent.  How 
much  was  realized? 


BANK  DISCOUNT. 


Banks  are  institutions  which  deal  in  money  and 
securities.  They  are  either  National,  State  or  Private 
banks. 

Their  principal  functions  are:  1st.  To  receive  de- 
posits ;  2d.  To  loan  money ;  3d.  To  these  is  added  a 
third  in  the  case  of  national  banks,  viz. :  To  issue  bank 
notes  which  pass  as  a  circulating  medium. 

The  law  does  not  prohibit  any  bank  from  issuing 
bills  if  it  chooses.  But  since  there  is  a  tax  of  10  per 
cent,  upon  the  circulation  of  all  banks,  except  those 
organized  under  the  provisions  of  the  National  Bank- 
ing act,  private  banks  are  practically  cut  off  from  this 
source  of  profit. 

National  banks  must  conform  to  certain  conditions 
imposed  by  Congress,  which  are  too  numerous  to  be 
mentioned  here.  By  complying  with  these  conditions 
they  are  granted  many  privileges  which  private  banks 
do  not  have. 

Banks  perform  a  very  useful  function  in  society.  It 
would  be  impossible  to  transact  the  enormous  volume 
of  business  in  the  country  if  money  were  used  in  every 
transaction.  It  is  estimated  that  95  per  cent,  of  the 
business  of  New  York  City  is  effected  by  exchange 


92  PRACTICAL  METHODS 

through  banks.  They  present  a  safe  form  of  deposit 
from  which  sums  of  money  may  be  checked  out  at 
any  time.  And  in  this  way  they  greatly  facilitate  local 
exchanges.  The  bank  is  the  purse  of  society  and 
readily  responds  to  every  demand  made  upon  it. 

The  manner  in  which  exchanges  are  made  between 
different  places  by  means  of  banks  will  be  explained 
under  the  head  of  exchange. 

Banks  are  the  usual  purchasers  of  notes,  and  they 
have  a  different  method  for  computing  discount  from 
the  method  given  under  True  Discount.  If  a  note  is 
due  in  three  months  the  bank  simply  subtracts  inter- 
est on  the  face  of  the  note  for  the  given  time. 

Bank  Discount  is  interest  on  the  face  or  amount  of 
the  debt  paid  in  advance. 

It  is  evident  that  by  this  method  the  discount  will 
be  greater  than  true  discount,  and  hence  the  rate  of 
profit  greater  to  the  creditor.  It  really  raises  the  rate 
of  interest  paid  by  the  debtor.  For  example,  a  man 
presents  his  note  at  the  bank  for  $100,  due  in  one 
year.  The  bank  discounts  it  at  10  per  cent.,  paying 
$90  in  cash.  At  the  end  of  the  year  the  bank  receives 
$100,  or  $10  interest.  Only  $90  was  loaned  by  the 
bank.  Hence  its  rate  of  interest  is  |^=11^  per  cent. 

EXAMPLE. — Find  the  bank  discount  and  proceeds  of 
a  note  for  $1,140  due  in  3  yr.  6  m.  21  da.,  the  rate 
of  discount  being  8  per  cent. 

Solution:  Int.  on  $1  at  8  %  for  3  yr.  6  m.  21  da.= 
.284§.  $l,140X.284f=$324.52Dis.  Ans.  $1,140— 
$324.52-4815.48  Proceeds.  Ans. 


IN  ARITHMETIC.  93 

EXAMPLES. 

(247)  A  man  realized  from  a  sale  notes  amounting 
to  $900,  which  were  due  in  9  mo.,  interest  being  8  per 
cent.      He  had  them   cashed  on   the  day   they  were 
given  at  a  bank  at  10  per  cent.     What  did  he  realize? 

NOTE.— To  find  the  amount  on  which  the  discount  is  computed, 
proceed  as  in  second  illustrative  example  under  true  discount. 

Find  the  date  of  maturity,  days  to  run,  proceeds  and 
bank  discount  on  the  following  notes: 

(248)  $961.75.  BLOOMFIELD,  Aug.  29, 1881. 
Three  years  after  date  I  promise  to  pay  to  Nicholas  Nickleby, 

or  order,  nine  hundred   and   sixty -one  and  ^   dollars,  value 
received,  interest  at  7  per  cent.         MARTIN  CHUZZLEWIT. 

Discounted  March  30,  1883,  at  8  per  cent. 

( 249 )  $  178.80.  SAN  FKANCISCO,  CAL.,  May  15, 1888. 
Ninety  days  after  date  I  promise  to  pay  to  B.  L.  Smith,  or 

bearer,  the  sum  of  one  hundred  and  seventy-eight  and  -ffa  dollars, 
with  interest  at  10  per  cent.  J.  M.  HAWLEY. 

Discounted  July  5,  1888,  at  10  per  cent. 

(250)  $230.  HILLSBOBO,  IA.,  Nov.  20, 1883. 
One  year  after  date  I  promise  to  pay  to  the  order  of  John 

Carter  two  hundred  and  thirty  dollars.  B.  R.  VALE. 

Discounted  Jan.  1,  1884,  at  8  per  cent. 

A  bank  pays  out  money  to  a  depositor  only  upon 
presentation  of  a  certificate  of  deposit,  or  a  check. 

A  check  is  the  written  order  of  a  depositor  ordering 
the  bank  to  pay  money  to  some  certain  person. 


('HECK. 


No.  23.  QUINCY,  ILL.,  May  2,  1887. 

first  HatioRof  BQR^  0f  0uir2e$. 

Pay  to  J.   C.   Harkness  or  bearer 
Three  hundred  and  sixteen DOLLARS. 

$3164°.  THOMAS  HEDGE. 


NOTE.  —  The  bank  must  cash  this  check,  provided  Thos.  Hedge 
has  a  sufficient  amount  deposited. 

RULE. 

When  you  have  the  Face   of  a  note,  Multiply  \    ^ 
"       "       "Proceeds  "       "      Divide      \      y 

The  Proceeds  of  $1. 

When  you  have  the  Discount,  Divide  by  the  discount 
of  $1. 

EXAMPLE.  —  Find  the  face  of  a  60  da,   note  which 
yields  $632  when  discounted  at  6  per  cent. 

Solution:     Proceeds  of  $1  =.9895.     $632  -K9895  = 
$638.70.     Ans. 

EXAMPLE.  —  The  discount  on  a  4  mo.  note  at  8  per 
cent,  was  $15.65.     What  was  the  face  of  the  note? 

Solution:     Eule  II.     Discount  on  $1  for  4  m.  3  da. 
at  8    %  ==.027J.      $15.65  is   2.7J  %  of  what  sum? 
=  $572.56.     Ans. 


To  find  the  rate  of  interest  corresponding  to  any 
given  rate  of  discount,  the  method  usually  given  in 
arithmetics  is  long  and  intricate.  la  following  out  the 
purposes  of  this  little  book  this  operation  is  much 


IN  ARITHMETIC.  95 

simplified  and  shortened.  No  objection  can  surely  be 
brought  against  any  method  that  saves  time  and  labor 
without  sacrificing  principle.  The  student  will  also 
notice  a  similar  abbreviation  in  the  reverse  process  of 
finding  the  rate  of  discount,  corresponding  to  a  given 
rate  of  interest. 

Rule.  —  Divide  the  rate  of  discount  by  the  proceeds, 
of  $1  for  the  given  time  and  rate. 

EXAMPLE.  —  If  a  bank  discounts  a  60  day  note  at  10 
per  cent,  what  is  the  rate  of  interest? 

Solution:     Proceeds  of   $1  for   63  da.   at   10  %  = 
.9825.     .10-H.9825=  lO^Yy  %.     Ans. 

EXPLANATION.  —  Every  dollar  on  the  face  of  the  note 
brings,  when  discounted,  only  98^.  lOc.  is  therefore 
received  as  interest  on.  98^.  What  per  cent,  is  lOc.  of 
Kule  2.  lQ+9S  =  lQ  .  Ans. 


EXAMPLES. 

(251)  What  is  the  rate  of  interest:     When  a  90  da. 
note  is  discounted  at  6,  7,  8,  9,  10  per  cent.  ? 

(252)  When  a  30  da.  note  is  discounted  at  8,  9,  10, 
11,  12  percent.? 

(253)  When  a  note  due  in  1  yr.,  without  grace,  is 
discounted  at  ^,  §,  £,  1,  1^,  1^  per  cent,  a  month? 

To  find  the  rate  of  discount  corresponding  to  a 
given  rate  of  interest. 

Rule.  —  Divide  the  rate  of  interest  by  the  amount  of 
SI  for  the  given  time  and  rate. 

EXAMPLE.  —  At  what  rate  should  a  bank  discount  a 
60  da.  note  so  as  to  receive  just  10  per  cent,  interest? 


96  PRACTICAL  METHODS 

Solution:     Ami.  of  $1  at  10  %  for  63  da.  =  $1.0175. 


EXPLANATION.  —  Every  $1  loaned  for  63  da.  at  10 
per  cent,  will  amount  to  $1.0175  at  the  end  of  the 
time.  $1.0175  is  therefore  the  sum  discounted.  Since 
$1  is  the  amount  loaned  at  10  per  cent.,  the  interest 
will  be  lOc.  What,  therefore,  must  $1.0175  be  dis- 
counted at  (multiplied  by),  so  as  to  produce  the  lOc. 
interest?  10c.-r-  1.0175  =  9f|^  %.  Ana. 

EXAMPLES. 

What  rate  of  discount  should  a  bank  charge— 

(254)  On  a  30  da.  note,  to  receive  8,  9,  10,  12.  15, 
18  per  cent.  ? 

(255)  On  a  90  da.   note,  to  receive  8,  10,  12,  15, 
18  per  cent.  ? 

(256)  On  a  note  due  in  1  yr.,  without  grace,  to  re- 
ceive 6,  8,  9,  10,  12  per  cent.  ? 


EXCHANGE. 


Exchange  is  the  method  of  paying  debts  by  means 
of  Drafts,  or  Bills  of  Exchange,  or  by  a  transfer  of 
credits. 

If  every  transaction  in  the  country  were  effected  by 
means  of  money,  the  loss  resulting  therefrom  would 
be  very  great,  and  many  exchanges  would  necessarily 
be  prevented.  It  would  not  only  make  enormously  in- 
creased demands  on  the  safe  means  of  transmitting 
money,  but  would  result  in  vast  risks,  which  would  be 
a  constant  source  of  danger  and  annoyance.  Money 


JN  ARITHMETIC.  97 

once  lost  could  not  be  restored.  Transactions  would 
be  delayed  by  the  consequent  handling  and  convey- 
ance of  large  sums  of  coin,  while  persons  would  be 
required  to  keep  by  them  large  sums  to  discharge 
debts  which  are  constantly  accruing. 

By  the  present  machinery  of  the  business  world,  if 
a  party  in  Chicago  wishes  to  discharge  a  debt  in 
Boston,  it  is  done  by  sending  a  Bill  of  Exchange. 

A  Bill  of  Exchange  is  a  general  term,  including 
drafts,  checks,  and  bills  of  exchange  proper.  They 
are  written  orders  to  pay  money,  on  distant  persons  or 
companies. 

Domestic  Exchange  is  the  operation  of  exchange 
within  the  limits  of  a  country. 

Foreign  Exchange  is  the  payment  of  debts  by  the 
interchange  of  obligations  between  different  countries. 

Drafts  are  used  for  domestic  exchange;  Bills  of 
Exchange  are  used  for  foreign  exchange. 

When  a  bill  of  exchange  is  drawn  it  is  generally 
written  in  triplicate,  called  a  set  of  exchange.  The 
three  bills  are  identical,  except  that  they  are  numbered 
first,  second  and  third  of  exchange.  These  are  sent 
by  different  routes,  to  prevent  delay  by  accident  of  any 
kind.  When  one  of  them  has  been  accepted  or  paid, 
the  others  are  void. 

BILL  OF  EXCHANGE. 

,£2,500.  BOSTON,  MASS.,  Dec.  31, 1888. 

At  eight  of  this  First  of  Exchange  (second  and  third  of  same 
tenor  and  date  unpaid),  pay  to  L.  P.  Morton  &  Co.,  two  thousand 
five  hundred  Pounds  Sterling,  and  charge  the  same  to  our 
account. 

To  PALMER  BEOS.,  J.  S.  OGILVIE  &  CO. 

London,  England. 


98  PRACTICAL  METHODS 

DRAFT. 

$676.40.  DENVER,  COL.,  Aug.  3, 1884. 

At  sight  pay  to  J.  W.  Martindale,  or  order,  six  hundred  and 
seventy  six  and  ^s  dollars,  and  charge  the  same  to  our  account. 

To  THIRD  NATIONAL  BANK,  JOHN  I.  BROWN  &  SON. 

St.  Paul,  Minn. 

THE  DRAWER  of  a  draft  is  the  one  who  signs  it,  or 
the  one  who  orders  the  money  to  be  paid. 

THE  DRAWEE  of  a  draft  is  the  party  who  is  ordered 
to  pay  it. 

THE  PAYEE  is  the  one  to  whom  payment  is  ordered. 

Drafts  are  negotiable  under  the  same  conditions  as 
promissory  notes.  When  the  holder  of  a  draft  sells  it, 
he  must  indorse  it,  by  which  he  is  held  liable  for  its 
payment. 

A  SIGHT  DRAFT  is  an  order  for  the  payment  of 
money,  payable  at  sight,  or  when  presented  for  pay- 
ment. 

A  TIME  DRAFT  is  one  which  is  payable  after  a 
specified  time  has  elapsed  from  the  date  of  the  draft, 
or  from  sight. 

If  a  party  in  Bloomfield  owes  $500  in  Chicago,  he 
could  pay  it  by  registered  letter,  by  money  order,  or 
by  draft. 

If  he  wishes  to  purchase  a  draft  he  does  so  by  a 
bank  which,  by  having  funds  deposited  in  some  corre- 
spondent bank  in  Chicago,  is  enabled  to  draw  drafts 
payable  at  that  place.  The  draft  is  sent  to  the  party 
in  Chicago,  to  whom  the  debt  is  owing,  who  may  have 
it  cashed  at  any  bank  in  the  city.  In  this  way  the 
debt  is  conveniently  discharged  without  remitting 
money.  If  the  draft  is  lost  or  stolen  no  loss  occurs  to 


IN  ARITHMETIC.  99 

either  party,  except  time,  as  it  can  be  duplicated  and 
sent  again. 

The  bank  of  Bloomfield  must  always  have  funds  de- 
posited in  its  correspondent  bank  in  Chicago  or  its 
drafts  will  not  be  honored.  If  this  amount  becomes 
large  through  various  interchanges  in  financial  opera- 
tions, it  may  sell  drafts  on  its  correspondent  at  less 
than  their  face  value,  in  order  to  lessen  the  amount  by 
inducing  people  to  buy  drafts. 

The  draft  is  then  said  to  be  at  a  Discount. 

If  its  balance  of  deposits  becomes  small,  it  will 
charge  more  than  the  face  value  for  drafts,  in  order  to 
check  the  paying  out  of  this  fund,  which  would  eventu- 
ally necessitate  the  shipment  of  coin  to  take  its  place, 
an  expensive  method  of  preserving  the  proper  amount 
of  reserve  fund. 

In  this  case  the  draft  is  sold  at  a  Premium. 

Drafts  are  frequently  cashed  at  banks  other  than 
the  one  named  in  the  draft.  This  necessitates  a  set- 
tlement between  different  banks.  In  the  larger  cities 
this  settlement  is  effected  by  means  of  the  Clearing 
House.  At  the  close  of  business  hours  representatives 
of  the  various  banks  meet  with  their  separate  accounts 
against  each  other.  These  accounts  are  balanced,  and 
the  sum  due  from  any  bank  paid  into  the  clearing 
house  fund.  In  a  few  minutes  accounts  aggregating 
millions  of  dollars  are  thus  correctly  settled. 

Sometimes  the  draft  is  drawn  on  private  parties.  It 
is  usually  presented  before  due,  and  the  drawee, 
"accepts"  it  or  not,  which  will  be  determined  by  the 
credit  of  the  drawer  of  the  draft.  An  acceptance  is 
made  by  writing  across  the  face  of  the  draft  "Ac- 


100  PRACTICAL  METHODS 

cepted,"  together  with  the  date  and  signature  of  the 
drawee.  The  draft  then  becomes  the  promissory  note 
of  the  drawee. 

In  case  the  draft  is  not  accepted  the  payee  falls 
back  upon  the  drawer  for  the  amount. 

RULES  FOR  EXCHANGE. 

When  you  have  the  Face  of  a  draft,  Multiply  \  -o 
"       "       "       "    Cost      "      "      Divide      (    y" 

1+Premium, 

or 
1 — Discount 

When  time  is  an  element  substract  the  bank  dis- 
count on  $1  for  the  given  time,  before  multiplying  or 
dividing. 

EXAMPLE. —  What  is  the  cost  of  a  sight  draft  on 
Cincinnati  for  $150  at  1  per  cent.  Premium? 

Solution:  1+premium  =  1.01.  The  face  of  the 
draft  is  given ;  hence,  $150 X  1.01  ==  $151.50.  Ans. 

EXAMPLE. — What  is  the  face  of  a  sight  draft  on 
New  Orleans  which  cost  $3,200,  exchange  being  ^  per 
cent.  Discount? 

Solution:  1—  £  %  disc.  =  .995.  $3,200  =  cost; 
hence,  $3,200-^.995=  $3,216. 08.  Ans. 

EXAMPLE. — What  will  be  the  cost  of  a  60  da.  draft 
on  Louisville  for  $648.70  at  £  per  cent.  Premium, 
interest  at  6  per  cent  ? 

Solution:  l+.00f  =  1.0075.  Bank  disc,  on  $1  for 
63  days  at  6  %  =-.0105.  1.0075  —  .0105  =  .997. 
$648.70=Face;  hence,  $648. 70 x.997=$646.75.  Ans. 


IN  ARITHMETIC.  101 

EXAMPLE. — What  was  the  face  of  a  12  da.  draft  on 
New  York  at  1  per  cent.  Discount,  interest  at  6  per 
cent.,  which  cost  $500? 

Solution:  1 — .01  =  .99.  Bank  disc,  on  $1  for  15 
da.  at  6  %  ==.0025.  .99— .0025  =  .9875.  $500  = 
Cost;  hence,  $500-K9875  =  $506.33.  Ans. 

EXAMPLES. 

Find  the  cost  of — 

(257)  A  30  da.  draft  on  Chicago  for  $240,  exchange 
being  ^  per  cent.  Discount,  interest  at  6  per  cent. 

(258)  A  draft  for  $1,250    on    Buffalo,   exchange 
at  1  per  cent.  Discount. 

(259)  A  90  da.  draft  on  Philadelphia  for  $672.30, 
exchange  at  f  per  cent.  Premium,  interest  at  6  per  cent. 

(260)  A  man  purchased  a  car-load  of  wheat  (500 
bu. )  at  74|c.  per  bu.,  and  paid  for  it  by  a  draft  for  the 
amount,  on  St.  Louis,  at  1^  per  cent.  Premium.    What 
did  the  draft  cost? 

(261)  What  was  the  face  of  a  draft  costing  $200, 
exchange  being  1  per  cent.  Discount? 

(262)  Find  the  face  of  a  60  da.  draft  on  Denver, 
exchange  at  1  per  cent.  Premium,  and  costing  $1,480, 
interest  at  6  per  cent. 

(263)  A  draft  on  New  York  costing  $416.50,  was 
purchased  at  ^  per  cent.  Premium.      What  was  its 
face? 


102  PRACTICAL  METHODS 

HORNER'S    METHOD    FOB    THE    EX- 
TRACTION  OF   ROOTS. 


Cube  root  may  be  extracted  much  more  easily, 
especially  if  more  than  two  figures  are  required  in  the 
answer,  by  Homer's  method  for  the  extraction  of  roots, 
than  by  the  method  generally  given  in  arithmetics. 
If  this  method  be  taught  from  the  first,  pupils  will 
readily  grasp  it,  and  it  soon  becomes  easy  for  them  to 
use  it. 

In  addition  to  this,  Horner's  method  may  be  used 
for  the  extraction  of  roots  higher  than  the  third,  since 
it  will  apply  to  any  root  whatever. 

This  method  is  explained  below,  and  we  hope  its  use 
may  be  extended. 

EXAMPLE.  —  Find  the  cube  root  of  100. 

Operation  : 

00  100    I  4.641+     Ans. 
4                16  64 

4800  -^6000 
120      5556         33336 

126      634800 

2684000 


1380  64588800 

1384  72656000 

1388 
13920 

EXPLANATION. 

As  many  columns  are  formed  as  the  number  of  the 
root  to  be  extracted.  At  the  head  of  the  last  column 
is  placed  the  number  rf  which  we  wish  to  extract  the 
root,  and  ciphers  for  the  others.  The  first  figure  of 


IN  ARITHMETIC.  103 

the  root  is  then  found.  For  100,  the  nearest  cube 
root  was  4.  This  number  is  added  to  the  1st  column, 
multiplied  by  itself  and  added  to  the  2d  column, 
multiplied  by  itself  again  and  subtracted  from  the 
given  number. 

Commencing  at  the  left  add  the  root  to  the  1st  col., 
multiply  the  result  by  the  root  and  add  to  the  2d  col . 

Add  the  root  to  the  1st  col. 

Then  add  one  cipher  to  the  1st  col.,  2  to  the  2d,  and 
3  to  the  3d. 

Divide  the  2d  col.  into  the  3d  for  the  next  figure  of 
the  root,  making  allowance  for  the  sum  to  be  added  to 
the  trial  divisor. 

Add  this  2d  figure  of  the  root  to  the  1st  col.,  multi- 
ply the  result  by  it  and  add  to  the  2d  col.,  multiply 
this  result  by  it  and  substract  from  the  3d  col. 

Add  the  2d  figure  of  the  root  to  the  1st  col.,  multiply 
the  result  by  it  and  add  to  the  2d  col. 

Add  it  to  the  1st  col.,  stopping  one  place  sooner 
each  time. 

Bring  down  ciphers  and  proceed  as  before. 

EXAMPLE. — Find  the  5th  root  of  36. 

Operation  : 

000 
248 
4  12  32 

6  24  8 

8  ^.fyty       816 

J0.0  832 


2.04767+    Ans. 

0 
16 

80.00015 
83264 
86591 
8718 
877t 
X 
X 

36. 
32 
~1.  00000 
3  33056 

61026 

5.918 
5266 

652 
614 

38 


104  PRACTICAL  METHODS 

Here  five  columns  are  formed,  and  the  work  is 
shortened  as  follows: 

Instead  of  adding  ciphers  strike  out  one  figure  from 
the  4th  col.,  two  from  the  3d,  three  from  the  2d,  and 
four  (if  there  be  that  many)  from  the  1st,  then  divide 
as  before,  and  again,  when  the  time  comes  to  add 
ciphers,  strike  out  figures  as  before. 

In  this  way  the  work  may  be  much  shortened,  as  in 
the  example  given. 

EXAMPLES. 

(264)  Find  the  cube  root  of  19683. ;  of  2,803,221. ;  of 
500,  to  three  decimal  places;  of  59319;  of  5359375. 

(265)  Extract  the  fifth  root  of  7962624;  of  10; 
of  50. 


IN  ARITHMETIC.  105 


SOME  USEFUL  FACTS. 


The  following,  though  not  all  strictly  belonging  to 
mathematics,  may  be  taught  in  connection  with  some 
of  the  common  school  branches.  They  should  induce 
the  teacher  to  make  discoveries  and  calculations  for 
himself,  and  collect  bits  of  information  not  found  in 
text  books,  yet  useful  in  every  day  life. 


The  Specific  Gravity  of  a  body  is  the  weight  of  a 
body  in  air,  divided  by  the  difference  between  its  weight 
in  air  and  in  water. 


The  principle  of  Specific  Gravity  was  discovered  by 
Archimedes,  a  famous  philosopher  in  Sicily  2000 
years  ago. 


The  foundations  of  Geometry  were  laid  by  Euclid, 
an  ancient  mathematician. 


Arithmetic  is  a  modern  science.  The  Romans  did 
not  carry  their  computations  very  far  because  their 
system  of  notation  was  too  cumbersome  to  admit  of 
this.  Arithmetic  really  began  with  the  introduction 
of  the  Arabian  method  of  notation. 


106  PRACTICAL  METHODS 

Algebra  was  introduced  into  Europe  by  the  Arabs 
Also  the  figures  1,  2,  3,  etc.,  which  are  universally 
used  to-day. 

The  weight  of  the  earth  is  6  sextillion  tons. 


The  rotation  of  the  earth  on  its  axis  tends  to  hurl 
bodies  on  its  surface  off  into  space.  This  centrifugal 
force  increases  from  the  poles  to  the  equator.  At  the 
latter  place  it  diminishes  the  weight  of  substances 


If  the  earth  rotated  with  a  velocity  17  times  as  great 
as  at  present,  this  force  would  equal  gravity,  and 
bodies  on  the  earth's  surface  would  not  weigh  any- 
thing. 

When  the  earth  was  thrown  off  from  the  sun  the 
impelling  force  acted  at  a  point  26  miles  out  of  line 
with  the  center  of  the  earth,  in  order  to  give  it  a  rotary 
motion  once  in  24  hours. 


Every  particle  of  matter  attracts  every  other  particle 
in  proportion  to  the  mass,  and  inversely  proportional 
to  the  square  of  the  distance.  This  is  termed  the  law 
of  Gravitation,  discovered  by  Sir  Isaac  Newton. 


A  body  on  the  sun  would  weigh  28  times  as  much 
as  on  the  earth.  A  man  weighing  150  Ibs.  here  would 
weigh  over  two  tons  on  the  sun's  surface. 


IN  ARITHMETIC.  T07 

The  North  or  Pole  star  is  li  degrees  from  the  true 
pole  of  the  heavens,  which  is  continually  changing  on 
account  of  an  oscillatory  movement  of  the  earth's  axis. 


The  stars  are  composed  of  much  the  same  materials 
as  the  earth.  Their  composition  is  ascertained  by  the 
use  of  the  spectroscope. 


TABLE  OF  SPECIFIC  GRAVITIES. 

The  following  table  shows  the  weight  of  a  few  com- 
mon substances  compared  with  the  weight  of  an  equal 
bulk  of  water : 

FOB  GASES—         Air,  1. 

Hydrogen,  .069 

Nitrogen,  .972 

Oxygen,  1.106 

FOB  LIQUIDS  AND  SOLIDS — 

Water,  1. 
Sea  Water,      1.026 

Cork,  .24 

Brick,  1.9 

Ice,  .92 

Limestone,  2. 5 

Mercury,  13.6 

Milk,  1.032 

Glass,  3. 

Iron,  7.7 

Steel,  7.8 

Lead,  11.3 

Silver,  10.5 

Gold,  19.25 


108 


PRACTICAL  METHODS 


CLASSIFICATION  OF   FIGURES  USED  IN 
MENSITRATION. 

I  Scalene — no  two  sides  equal. 
Triangles:  <  Isosceles — two  sides  equal. 

(  Equilateral — All  three  sides  equal. 

(  Trapezium — no  two  sides  parallel. 
Quadrilaterals:  <  Trapezoid — two  sides  parallel. 

(  Parallelogram— opposite  sides  parallel. 

f  Rhomboid  —  a    parallelogram   with 

obligue  angles. 

Rhomb — an  equilateral  rhomboid. 
Rectangle  —  a    parallelogram    with 

right  angles. 
Square — an  equilateral  rectangle. 


Parallelograms :  -| 


Solids:-! 


Cube  —  a   solid   whose   faces    are   6    equal 

parallelograms. 
Globe  —  a  solid  whose  surface  is  everywhere 

equally  distant  from  the  center. 
Pyramid   —  a   solid   with   triangular    faces 

meeting  at  the  apex. 
Cone  — a  pyramid  with  an  infinite  number 

of  sides. 


RULES   FOR   MEASUREMENT   OF   FIGURES. 

To  find  the  area  of  a  triangle: 

Base  x  ^  Altitude. 

NOTE. — The  altitude  of  a  triangle  is  the  perpendicular  distance 
from  the  base  to  the  vertex  of  the  opposite  angle. 


IN  ARITHMETIC.  109 

To  find  the  area  of  a  trapezium: 
Divide  by  diagonals  the  figure  into  triangles.     Take 
the  dimensions  of  each  and  solve  by  rule  for  triangles. 


To  find  the  area  of  a  trapezoid : 

Multiply  half  the  sum  of  the  parallel  sides  by  the 
perpendicular  distance  between  them. 


To  find  the  area  of  a  parallelogram : 
Multiply  any  side  by  the  perpendicular  distance  be- 
tween it  and  its  opposite  side. 


To  find  the  surface  of  a  cube: 

Square  one  side  of  the  cube  and  multiply  by  6. 


To  find  the  surface  of  a  globe: 
Multiply  the  diameter  by  the  circumference,  or  mul- 
tiply the  Di.s  by  8.1416. 


To  find  the  surface  of  a  pyramid: 

Find  by  a  preceding  rule  the  area  of  each  face  and 
multiply  by  the  number  of  faces.  To  this  add  the 
area  of  the  base. 


To  find  the  surface  of  a  cone: 

Multiply  the  circumference  of  the  base  by  ^  the 
slant  height. 


110  PRACTICAL  METHODS 

To  find  the  solidity  of  a  cube: 
Cube  the  length  of  one  side. 


To  find  the  solidity  of  a  globe: 

Cube  the  diameter  and  multiply  by  .5236. 


To  find  the  solidity  of  a  pyramid : 

Multiply  the  area  of  the  base  by  |  the  altitude. 


To  find  the  solidity  of  a  cone: 

Multiply  the  area  of  the  base  by  |  the  altitude. 

NOTE. — The  altitude  of  a  pyramid  or  cone  is  the  perpendicular 
distance  from  base  to  apex. 


PEACTICAL   PKOBLEMS. 

No  answers  will  be  given  to  the  following  problems, 
which  every  teacher  should  be  able  to  work.  They 
are  given  as  suggestions  as  to  the  kind  of  examples 
which  the  teacher  should  give  his  pupils. 

I.  How  many  tons  of  ice  in  an  icehouse  22x1(3,  the 
ico  being  packed  in  solid  to  a  depth  of  12  ft.  ? 

II.  Find  the  cost  of  a  board  sidewalk  4  ft.  wide 
from  the  door  of  the  school-house  to  the  street,  the 
price  of  lumber  being  given. 

III.  How  many  scholars  are  present  to-day  ?    What 
per  cent,  is  that  of  the  whole  number  enrolled? 


IN  ARITHMETIC.  Ill 

IV.  Write  a  note  payable  to  the  order  of  some  one, 
and  have  it  properly  endorsed. 

V.  Write  a  receipt,  a  due  bill,  a  check,  a  draft,  a 
note  with  security. 

VI.  What  is  the  total  cost  price  each  year  of  all 
the  books  used  in  this  school,  and  how  much  money 
must  be  placed  on  interest  at  8  %  to  meet  this  expense  ? 

VII.  Find  out  by  a  careful  calculation  which  would 
be  the  most  profitable ;  to  invest  your  money  in  a  farm, 
or  loan  it  at  4  %  ? 

VIII.  Find  out  what  the  taxes,  interest  on  money 
invested,  repairs,  insurance,  etc.,  on  your  house  are 
annually,    and   then   ascertain   whether   it    would    be 
cheaper  to  rent  or  own  property. 

IX.  How  many  cords  of  wood  can  be  piled  in  a 
wood  shed  15  ft.  long,  10  ft.  wide,  6  ft.  high? 

X.  What  does  a  box  of  water  weigh,  3  ft.  each  way  ? 
NOTE. — A  cubic  foot  of  water  weighs  62^  Ibs. 

XI.  Calculate   the   weight   of   the   earth,    specific 
gravity  5§. 

XII.  How  many  freight  cars  would  it  take  to  haul 
the  grain  that  could  be  put  in  your  school-room  ? 

XIII.  If  a  box  contain  3  bu.  of  potatoes,  iiow  much 
would  a  box  hold  twice  as  large  each  way  ? 

XIV.  How  many  sq.  rods  in  your  school  yard? 
How  many  in  a  lot  3  times  as  large  each  way  ? 

XV.  Two  notes  for  $100  each  are  discounted,  one 
for  30  da.,  the  other  for  90  da.     Which  would  yield 
the  greatest  rate  of  interest  to  the  bank  ? 


112  IN  ARITHMETIC. 


ANSWERS. 


1. 

7,221. 

29. 

1,111|. 

2. 

9,009. 

30. 

16,044f. 

3. 

616. 

31. 

23,51H. 

4. 

13,224. 

32. 

7,511-}. 

5. 

2,516. 

33. 

3,690. 

6. 

2,464. 

34. 

1,728. 

7. 

1,649. 

35. 

2,988. 

8. 

2,925. 

36. 

92,664. 

9. 

9,120. 

37. 

615,450. 

10. 

8,096. 

38. 

4,449,955. 

11. 

985,050. 

39. 

57,446.4  gal. 

12. 

977,112. 

40. 

2,225.3  gal. 

13. 

9,000. 

'41. 

166J  bbl. 

14. 

115,200. 

42. 

12,927.6  bbl. 

15. 

4,713|. 

43. 

3,776  gal. 

16. 

7,950. 

44. 

47,000  gal. 

17. 

3,393,333|. 

45. 

1,475  gal. 

18. 

26,992. 

46. 

81  bbl. 

19. 

6,241. 

47. 

140f  bbl. 

20. 

3,721. 

48. 

2,856.7  gal. 

21. 

2,304, 

49. 

2,127.3  gal. 

22. 

1,089. 

66.9  bbl. 

23. 

10,609. 

50. 

48  bu.     24  bu. 

24. 

2,704. 

51. 

429  bu.     858  bu. 

25. 

729. 

52. 

6,350.4  bu. 

26. 

9,801. 

53. 

7.26  tons. 

27. 

8,7111* 

54. 

13  tons. 

28. 

9,3444. 

55. 

11  —  tons. 

PRACTICAL  METHODS  1U 

56.  2.28  tons.  81.  28^V 

57.  1.55  tons.  82.  61^- 

58.  1.77  tons.  83. 

59.  41+  perch.  84. 

60.  13|  perch.  85. 

61.  14|  perch.  86. 

62.  1,750  brick.  87. 

63.  1,950  brick.  88. 

64.  240  gal.  89.  ft. 

65.  419,904  gal.  90.  1. 

66.  40,500  gal.  91.  80. 

67.  66|  ft.     104^  ft.  92.  1,880. 
266|  ft.     6,666§  ft.       93.  32  jV 

68.  28.8  mi.      22—  mi.       94.  947\. 
178  mi.  95.  452f|. 

69.  400ft.     1,600ft.  96.  416§. 
14,400ft.    57,600ft.       97.  22|. 

70.  l|sec.     2Jsec.  98.  3T90. 
12|  sec.     3.95  sec.  99.  24||. 

71.  112ft.  100.  3££. 

72.  900ft.  101.  T5T. 

73.  1,602  ft.     $23.59.  102.  8|f 

74.  8.4  in.     11.2  in.  103.  trAV 
16.8  in.  104.  T\. 

75.  50  ft.     66§  ft.  105.  25.45. 
91§  ft.  106.  74.8. 

76.  51.84  bunches.  107.  2.1445. 
39  j  bunches.  108.  1616.588. 

77.  30T7T  bunches.  109.  $52,465.536. 

78.  66|  bunches.  110.  $60.24. 

79.  15T£.  111.  .521. 

80.  103T8A5r-  112.  6.44. 


114 


PRACTICAL  METHODS 


113.  12.9. 

114.  5.8296. 

115.  38.506. 

116.  $532.82. 

117.  The  N.  E.  J  of  the 
S.  E.  J  of  Section 
19,  Twp.  3  North, 
Range    3    West   of 
the  5th  P.  M. 

The  W.  £  of  the  N. 
W.  JoftheN.W.J 
of  Section  10,  Twp. 
4  North,  Range  1 
West  of  the  5th  P. 
M.,  containing  20  A. 
The  South  £  of  the 
N.  E.  4  of  the  N. 
W.  ^  of  Section  33, 
Twp.  2  North, 
Range  6  West,  con- 
taining 20  A. 
The  N.  E.  J  of  the 
N.  E.  l  of  Section 
22,  Twp.  4  North, 
Range  5  West,  con- 
taining 40  A. 

118.  80  A. 

119.  20  A.     80  A. 

121.  3hr.  13m.  49f  sec. 
P.  M. 

122.  5hr.  44m.  26f|sec. 
P.  M.  of  the  previ- 
ous day. 


123. 

2°  20'  East. 

124. 

90°  12'  15"  West. 

125. 

7hr.  20  m.  56  sec. 

j 

A.  M.,  Tuesday. 

126. 

E.  13°  23'. 

127. 

6  hr.  51  m.  12  sec. 

» 

A.  M.  of  the  follow- 

ing  day. 

128. 

12. 

129. 

4.^ 

130. 

TV 

131. 

5^. 

132. 

9 

¥¥' 

133. 

4 

T- 

134. 

Hi- 

135. 

-III- 

136. 

$18.90. 

137. 

25,298  ft. 

138. 

67H  ft. 

139. 

34357.  86  mi. 

140. 

1    663  1  ftA/~* 

141. 

5    pipes. 

142. 

$230.11T4T. 

143. 

4|f  da. 

144. 

$5,222.40. 

145. 

2i-|4|  yds. 

146. 

435  HI  ft. 

147. 

11^  hrs. 

148. 

$470. 

149. 

64}. 

150. 

VT- 

151. 

200. 

153. 

208.35. 

IN  ARITHMETIC. 


115 


154. 

.065|. 

155. 

156  cattle. 

156. 

lOff. 

157. 

A*- 

158. 

A*- 

159. 

88f%. 

160. 

Q  2  9  of 

161. 

%^-tf 

162. 

4|%. 

163. 

800. 

164. 

$1.40f. 

165. 

5,520  men. 

166. 

31*  gal. 

167. 

4^-  mi. 

168. 

$75. 

169. 

35  horses. 

170. 

25%  greater. 

20%  less. 

171. 

21i|  %  greater. 

17  1%  less. 

172. 

68|  %  greater. 

40ff%  less. 

173. 

61^f-%  greater. 

38^%  less. 

174. 

Com.  $129.87. 

Pro.  $3,116.88. 

175. 

$73.96. 

176. 

$39. 

177. 

i%. 

178. 

3Ji%. 

179. 

22T%' 

180. 

10  bbls. 

181. 

$884.21. 

182. 

1st  com.,  $36. 

2d  com.,  $33.23. 

Sugar,  $830.77. 

183. 

$1,435.     $1,358. 

184. 

$321.43. 

185. 

$103.50. 

186. 

14|%. 

187. 

?»* 

188. 

$8,500. 

189. 

$65.80. 

190. 

$1,200. 

191. 

$51.20. 

192. 

6ff|%. 

193. 

8-^%. 

194. 

$69.75. 

195. 

$50.62J. 

196. 

4%. 

197. 

5^%. 

198. 

39   sharea 

199. 

105    shares. 

200. 

$700, 

201. 

500  shares. 

202. 

$1,503.75. 

203. 

$428.875. 

Sell'g  pr.,$2,865.25 

204. 

1J%. 

205. 

$3,000. 

206. 

$2,000. 

207. 

$850. 

208. 

$1,000. 

116 


PRACTICAL  METHODS 


209. 

$1,050. 

210. 

Denver&RioGrande 

239. 

Stock. 

240. 

211. 

$7,704. 

212. 

$6,012.50. 

213. 

<** 

241. 

214. 

w.<- 

242. 

215. 

86f%. 

243. 

216. 

80%. 

244. 

217. 

$44.82. 

245. 

218. 

$41.64 

246. 

219. 

$155.50. 

247. 

220. 

$194.61. 

248. 

221. 

$300.07. 

222. 

1  yr.  3  m. 

223. 

2  yr.  2  m.  7  da. 

224. 

1  yr.  6  m.  6  da. 

249. 

225. 

7%. 

226. 

7%. 

227. 

8%. 

228. 

$2,507. 

229. 

$672. 

230. 

$732.82. 

250. 

231. 

$600. 

232. 

$479.80. 

233. 

$412.50. 

234. 

$2,210.75. 

251. 

235. 

$72.69. 

236. 

$7.41. 

237. 

$80.38. 

252. 

238. 

Date   of    Maturity, 

Sept.  7,  1879. 

Amount,  $193.60. 

Amount  $689.87. 
Date    of     Maturity, 
May  10—  13,  1884. 
Amount,  $2,744.92. 
$1,221.89. 
$10.00. 
$554.03. 

$1,201.96.      $9,97. 
$387.72. 
$1,299.93. 
$882.21. 

Date    of    Maturity, 
Sept.  1,  '84. 
Proceeds,  $1032.07. 
Discount,  $132.21. 
Date   of     Maturity, 
July  13—16,  1888. 
Time  to  run,  42  da. 
Proceeds,  $181.28. 
Discount,  $2.14. 

Due,  Nov.  23,  1884. 
Time  to  run,  327  da. 
Proceeds,  $213.29. 
Discount,  $16.71. 


2  9  7    r£  . 
"O  6  T  ?0  , 


ni  33  1 


IN  ARITHMETIC.  117 

253.     6|$%  I  8£U;  9||%;     257.     $237.48. 
k;  17-B%;         258.     $1,237.60. 


254. 

26°- 


.     261-     ^202.02. 
255.     7|fff%;  9TVfl%;     262.     $1,480.74. 

5     26a 


sr  (T  Q  "q" 


-  (    Q  q         • 

256.     5ff%;7^7%;  264'  27;  141;  7.934;  39; 

;9TV%;  175' 

265.  24;  1,585;  2.187. 


/AT    TEACHING    PATRIOTISM    YOU    WILL    BE     GREATLY     ASSISTED 
KNOWING  //Olf   THE  LAM'S  ARE  MADE  A.\I>  EXECCTJJ  >. 
THERE  IS  .\O  fiOOK  ISSTEI)    TO  COMPARE 
I.\  THIS  RESPECT  If/ Til 

HISTORY  AND    .    .    .  ^_ 

CIVIL  GOVERNMENT 


OF 


IOWA. 


For  the    Use  of  Normal  and  Public  Schools,    Teachers'  Institutes 
and  Private  Instruction. 

By  GEORGE  CHANDLER, 

Superintendent  of  Schools,  Osage,  lo-wa. 

The  book  is  revised  to  date  and  contains : 

HISTORY  OF  IOWA.  It  gives  a  brief  history  of  the  settlement  and  growth  of  the 
State.  No  resident  of  the  State  should  neglect  reading  and  studying  this  chapter.  It  is 
highly  interesting. 

INSTITUTIONS  OF  IOWA.  An  interesting  account  of  the  various  State  insti- 
tutions is  given;  when  established,  where  located  now,  how  supported,  how  governed,  in- 
mates, etc.  This  chapter  is  worth  the  price  of  the  book. 

CONSTITUTION  OF  IOWA,  The  Constitution  is  given  in  full  with  side  notes 
for  each  article.  Any  item  can  be  readily  found  by  running  the- eye  down  the  side  notes. 
This  arrangement  will  save  much  time,  and  is  a  valuable  feature  of  the  book. 

OFFICERS  OF  THE  STATE.  A  full  list  of  the  various  State,  District,  County, 
Town,  and  Township  Officers,  and  how  elected,  is  given.  Also  the  duties  of  each,  the 
amount  of  bond  required,  the  salary  received,  and  other  information  pertaining  to  each. 
No  intelligent  citizen  should  be  ignorant  of  these,  yet  I  fear  that  there  aie  many  boys  and 
girls  graduated  every  year  with  a  smattering  of  Greek  and  I^atin,  that  cannot  name  and 
give  duties  of  the  State  officers,  and  perhaps  there  are  a  few  teachers  likewise  deficient. 

OUTI/INES.  Several  pages  of  outlines  suitable  for  Class  Room  use  are  given  and 
will  be  found  valuable  in  Institute  work. 

REVISED  TO  DATE.  The  book  is  up  with  the  times,  containing  all  of  the  latest 
laws,  changes  in  salaries  and  duties  of  various  officers. 

ADOPTED  FOR  EXCI/USIVE  USB  in  several  counties  by  the  County  School 
Boards  for  terms  of  five  years.  This  shows  its  value  as  a  text  book. 

That  the  book  supplies  a  long  felt  want  no  one  can  deny.  It  is  written  in  a  pleasant 
forcible  manner.  It  has  already  been  adopted  in  many  schools  and  is  a  regular  text  book 
in  several  County  Institutes.  Every  parent,  every  pupil,  every  teacher,  should  have  a  copy. 

BOUND  IN  CLOTH,  170  PAGES.     MAILING  PRICE,  60  CENTS. 

A  very  low  price  will  be  made  for  Introduction,  Institute  Use,  and  to  Agents. 

A.  RLANAQAN,  Chicago. 


UNIVERSITY  OF  CALIFORNIA  AT  LOS  ANGELES 

THE  UNIVERSITY  LIBRARY 
This  book  is  DUE  on  the  last  date  stamped  below 


JIM 


>9    1940 


M>& 

wrrnwii 

OCT 


N       3 


Form  L-8 
20m-12,'3»(3as8) 


FEB 1 4  1952 


MAR  t 


7953 


SEP  1  7  1953 


;i€ 


A     000933214     9 


103 

G13p 

1892 


:ALli< 


BRANCA 

CALIFORfifc 


LOS   ANGELES,  CALIF. 


